A ball sliding on a rail: Conservation of energy and centripetal force

Question:
A ball is released at a certain height to slide on the rail as shown. It goes through the straight and circular track that is ended at the highest position C. The radius of the circular rail is $R$. The angle $\theta$ is taken from C to position of the ball. The mass and velocity of the ball are denoted as $m$ and $v$, respectively. We assume that there is no air resistance.
(1) What is the normal force so that the ball keeps sliding on the rail?
(2) Use conservation of energy. Find the angle when the ball gets off from the track.
(3) What is the minimum height to release to reach the maximum point C?

Answer:
(1) When the normal force, gravitational force, and centripetal force are balanced, the ball barely tracks the rail. From the Newton's equation of motion in a circular track is given as
$$\begin{equation} \sum F = N + mg\cos\theta = \frac{mv^2}{r} \end{equation}$$
Note that the direction toward center is defined as positive. Thus, we have
$$\begin{equation} N = \frac{mv^2}{R} - mg\cos\theta \end{equation}$$

(2) The height of the ball in the circular track can be expressed as $R+R\cos\theta$. From conservation of energy, the initial total energy must be equal to the final total energy.
$$\begin{equation} mgh = mgR(1+\cos\theta) + \frac{1}{2}mv^2 \end{equation}$$
From (2) and (3), eliminate $v$.
$$\begin{equation} N = mg \left( \frac{2h}{R} -2 -3\cos\theta \right) \end{equation}$$
When $N=0$, the ball gets off from the track. Substitute zero into (4) and solve for $\theta$.
$$\begin{eqnarray} & & \cos\theta = \frac{2}{3}\left(\frac{h}{R}-1\right)  \\ &\rightarrow& \theta = \cos^{-1} \left[\frac{2}{3}\left(\frac{h}{R}-1\right) \right] \end{eqnarray}$$

(3) When $\theta$ is zero in (5), the ball falls off at C. Since $\cos 0 = 1$, the minimum height $h$ is
$$\begin{equation} h = \frac{5}{2}R \end{equation}$$

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Rotational motion and tension on strings

Question:
Consider a rod AB with a horizontal arm AC. The length of AC is $a$ and strings are attached from A and C to hang a mass at D. The lengths of the strings are $a$ as shown.
(1) Find the tension on the string AD.
(2) Rotate the rod AB and the mass obtains a constant circular velocity, $v$. There is no slack with string CD. Find the tensions on strings AD and CD.
(3) Find the velocity and angular velocity of the object if string CD does not get slack.

Answer:
(1) The tensions on AD and CD are equal and balanced with the gravitational force, $mg$. Thus we have the tension as follows:
$$2T\cos 30^o = mg  \\ \rightarrow T = \frac{\sqrt{3}}{3}mg$$

(2) The radius of rotation is $\frac{a}{2}$, so the centripetal force is $\frac{mv^2}{a/2}$. Then, write out the equations in $x$- and $y$-axes.
$$\begin{eqnarray*} \mbox{For x, } \quad T_1\sin 30^o &=& T_2\sin 30^o + \frac{mv^2}{a/2}  \\ \mbox{For y, } \quad T_1\cos 30^o &+& T_2\cos 30^o = mg \end{eqnarray*}$$
Solving the simultaneous equations, we have
$$T_1 = m\left(\frac{\sqrt{3}g}{3}+\frac{2v^2}{a} \right)  \\ T_2 = m\left(\frac{\sqrt{3}g}{3}-\frac{2v^2}{a} \right)$$

(3) If the string does not get slack, the tension must be equal to or greater than zero. Let $T_2 \geq 0$.
$$m\left(\frac{\sqrt{3}g}{3}-\frac{2v^2}{a} \right) \geq 0  \\ \rightarrow \frac{\sqrt{3}g}{3}-\frac{2v^2}{a} \geq 0  \\ \rightarrow \frac{2v^2}{a} \leq \frac{\sqrt{3}g}{3}  \\ \rightarrow v \leq \sqrt{\frac{\sqrt{3}ga}{6}}$$
Since angular velocity, $\omega$, is velocity divided by radius. we obtain
$$\omega = \frac{v}{a/2} \leq \sqrt{\frac{2\sqrt{3}g}{3a}}$$

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Projectile onto an inclined plane: The maximum range

Question:
A projectile is launched from the origin with initial velocity $v_0$ and the angle $\theta$ from the $x$-axis. The object is reached on the inclined plane below $x$-axis with angle of $\alpha$. Assume that the mass, initial velocity and angle $\alpha$ are fixed. Then, what is the angle $\theta$ to obtain maximum range on the inclined plane?

Answer:
The position of the projectile is given as
$$\begin{eqnarray} x &=& v_0\cos\theta \ t  \\ y &=& v_0\sin\theta \ t - \frac{1}{2}gt^2 \end{eqnarray}$$
For the inclined plane, we can have the relationship between $x$ and $y$ as follows:
$$\begin{equation} y = -x\tan\alpha \end{equation}$$
Use (3) to eliminate $y$.
$$\begin{equation} -x\tan\alpha = v_0 \sin\theta \ t - \frac{1}{2}gt^2 \end{equation}$$
From (1), we have $t = \frac{x}{v_0 \cos\theta}$ to eliminate $t$.
$$\begin{eqnarray} -x\tan\alpha = v_0 \sin\theta\frac{x}{v_0 \cos\theta} - \frac{1}{2}g\left(\frac{x}{v_0 \cos\theta}\right)^2   \\ -x\tan\alpha = x\tan\theta - \frac{1}{2}\frac{gx^2}{v_0^2 \cos^2\theta} \\ x(\tan\alpha + \tan\theta) - \frac{1}{2}\frac{gx^2}{v_0^2 \cos^2\theta} = 0  \\ x \left\{ (\tan\alpha + \tan\theta) - \frac{1}{2}\frac{gx}{v_0^2 \cos^2\theta} \right\} = 0 \end{eqnarray}$$
$x=0$ is also the solution, but we are interested in the other one. Now, we can solve for $x$ as follows:
$$\begin{eqnarray} x &=& \frac{2v_0^2\cos^2\theta (\tan\alpha + \tan\theta)}{g}\\   &=& \frac{2v_0^2\cos^2\theta \left(\frac{\sin\alpha}{\cos\alpha} + \frac{\sin\theta}{\cos\theta} \right)}{g}\frac{\cos\alpha}{\cos\alpha}\\  &=& \frac{2v_0^2\cos\theta (\sin\alpha\cos\theta + \sin\theta\cos\alpha)}{g\cos\alpha} \\  &=& \frac{2v_0^2\cos\theta \sin(\alpha+\theta)}{g\cos\alpha} \\  &=& \frac{2v_0^2 \left\{ \frac{1}{2}\sin(2\theta+\alpha)+\sin\alpha\right\}}{g\cos\alpha} \\  &=& \frac{v_0^2 \left\{\sin(2\theta+\alpha)+\sin\alpha\right\}}{g\cos\alpha} \end{eqnarray}$$
$v_0$ and $\alpha$ are constant, so the value of $\sin(2\theta+\alpha)$ affects the value of the whole function. A sine or cosine function oscillates between -1 and 1 assuming that the amplitude is 1. Thus, when we have
$$\begin{equation} \sin(2\theta+\alpha) = 1 \end{equation}$$
the range becomes maximum. The angle must be
$$\begin{eqnarray} 2\theta+\alpha &=& \sin^{-1}1 = \frac{\pi}{2} \\ 2\theta &=& \frac{\pi}{2}-\alpha  \\ \theta &=& \frac{\pi}{4} - \frac{\alpha}{2} \end{eqnarray}$$

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A projectile motion: Bouncing back from a wall

Question:
As shown in the figure, an object is launched at initial velocity $v_0$ with the angle of $\theta$ toward the wall. The distance between A and B is $R$. The wall is frictionless and the coefficient of restitution with the object is $e$. The initially launched object bounces at C and comes back to the original position A through the projectile D. Assume that there is no air resistance. Answer the following questions:
(1) Find the time $t_1$ from A to C.
(2) Find the height $h_1$.
(3) After colliding the wall, find the time $t_2$ when reaching the highest point in the projectile D.
(4) Find the height $h_2$.
(5) Find the time $t_3$ from the highest position of projectile D to A.
(6) Find the coefficient of restitution $e$.

Answer:
(1) The initial velocity in the horizontal direction is expressed as $v_0\cos\theta$, which is constant through the motion. Therefore, the range $R$ is expressed by $v_0\cos\theta \times t_1$, so
$$t_1 = \frac{R}{v_0\cos\theta}$$

(2) Consider the vertical direction. The initial velocity is $v_0\sin\theta$ and the time is given in the above.
$$\begin{eqnarray*} h_1 &=& v_0\sin\theta t_1 - \frac{1}{2}gt_1^2  \\   &=& v_0\sin\theta \frac{R}{v_0\cos\theta} - \frac{1}{2}g\left(\frac{R}{v_0\cos\theta}\right)^2\\   &=& R\tan\theta - \frac{1}{2}g\left(\frac{R}{v_0\cos\theta}\right)^2 \end{eqnarray*}$$

(3) Let's consider the motion from A to the highest point of D. The time is defined as $t'$. Use the kinematic equation.
$$v_y = v_0\sin\theta -gt'$$
Since $v_y=0$ at the highest point, we obtain $t'$ as follows.
$$gt' = v_0\sin\theta \quad \rightarrow \quad t' = \frac{v_0\sin\theta}{g}$$
However, $t_2$ is from C, so subtract $t_1$ from $t'$.
$$t_2 = t' - t_1  = \frac{v_0\sin\theta}{g} - \frac{R}{v_0\cos\theta}$$

(4) The height $h_2$ is given by the kinematic equation.
$$h_2 = v_0\sin\theta t' - \frac{1}{2}gt'^2 \\ = \frac{(v_0\sin\theta)^2}{g}-\frac{1}{2}\left(\frac{v_0\sin\theta}{g}\right) \\ =  \frac{v_0\sin\theta (2v_0\sin\theta-1)}{2g}$$

(5) The motion is symmetric, so
$$t_3 = t'$$

(6) Define the coefficient of restitution by the distances before and after collision; namely, we have
$$e = \frac{\mbox{distance before collision}}{\mbox{distance after collision}}$$
The distance before collision is $R$. The distance after collision is that the total time, $t_2+t_3$, times velocity $v_0\cos\theta$. Thus,
$$e=\frac{R}{v_0\cos\theta \times (t_2 + t_3)}  \\  = \frac{R}{v_0\cos\theta (2t' - t_1)}  \\  = \frac{R}{\left( \frac{2v_0 \sin\theta}{g}-\frac{R}{v_0 \cos\theta}\right) v_0 \cos\theta}  \\  = \frac{R}{\frac{2v_0^2 \sin\theta\cos\theta}{g}-R}  \\  = \frac{R}{\frac{2v_0^2 \sin\theta\cos\theta-gR}{g}}  \\  = \frac{gR}{2v_0^2 \sin\theta\cos\theta-gR}$$

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Motion of an object inside an open box

Question:
As shown in the figure, an open box is put on frictionless floor; then, an object is placed in the middle of the box. There is no friction between the object and the surface of the box. The masses of the object and the box are equal. The coefficient of restitution between the object and inside box is $e \ (0<e<1)$.
(1) Let the object move at $v_i$ in the positive $x$ direction. Find the velocity of the object related to the floor after colliding the inside box.
(2) What is the position of the middle of the box after the second collision in the other side?

Answer:
(1) There is no external force; therefore, we can consider conservation of momentum. Let $v$ and $u$ be the velocities of the object and the box, respectively. The subscripts, $i$ and $f$, represent initial and final. Thus, we have
$$\begin{equation} mv_i = mv_f + mu_f \end{equation}$$
The coefficient of restitution is defined as the relative velocity after collision divided by the relative velocity before collision. Namely,
$$\begin{equation} e = \frac{v_f - u_f}{0 - v_i} \end{equation}$$
From (1) and (2), we have
$$\begin{eqnarray} v_i &=& v_f + u_f \\ -v_i e &=& v_f - u_f \end{eqnarray}$$
Let's find $v_f$ in terms of $v_i$ by eliminating $u_f$.
$$\begin{equation} v_f = \frac{1-e}{2}v_i \end{equation}$$

(2) First, we find the final velocity of the box by eliminating $v_f$ from (3) and (4).
$$\begin{equation} u_f = \frac{1+e}{2}v_i \end{equation}$$
The time from the first to the second collision is the distance, $L$, divided by the relative velocities.
$$\begin{equation} t = \frac{L}{u_f - v_f} = \frac{L}{ev_i} \end{equation}$$
Therefore, the distance that the box moved is
$$d = t \times u_f = \frac{L}{ev_i} \times \frac{1+e}{2}v_i = \frac{1+e}{2e}L$$

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Differential equation: A special case ($f'(x) - f(x) =...$)

Question:
Solve a differential equation.
$$f'(x) - f(x) = -2\sin x$$
The initial condition is $f(0) = 1$.

Answer:
When you find the form, $f'(x) - f(x) =...$, we can utilize the following relation:
$$[f(x)e^{-x}]' = e^{-x}[f'(x) - f(x)]$$
If we apply this formula to the above differential equation, we get
$$[f(x)e^{-x}]' = e^{-x}[-2\sin x]$$
For the right hand side, we used $f'(x) - f(x) = -2\sin x$. Then, integrate both sides of above equation.
$$\begin{equation} f(x)e^{-x} = -2\int e^{-x}\sin x dx \end{equation}$$
For the right hand side, we just consider integration by parts of the following factor:
$$\begin{eqnarray*} & & \int e^{-x}\sin x dx = -e^{-x}\sin x - \int (-1)e^{-x}\cos x dx  \\ \rightarrow & & \int e^{-x}\sin x dx = -e^{-x}\sin x - e^{-x}\cos x - \int (-1)e^{-x}(-1)\sin x dx  \\ \rightarrow & & \int e^{-x}\sin x dx + \int e^{-x}\sin x dx = -e^{-x}\sin x - e^{-x}\cos x   \\ \rightarrow & & 2\int e^{-x}\sin x dx = -e^{-x}(\sin x + \cos x)   \\ \rightarrow & & \int e^{-x}\sin x dx = -\frac{1}{2}e^{-x}(\sin x + \cos x) + C \end{eqnarray*}$$
Plug this into (1).
$$\begin{eqnarray} f(x)e^{-x} &=& e^{-x}(\sin x + \cos x) + C  \\ \end{eqnarray}$$
Use the initial condition, $f(0) = 1$.
$$f(0) = \sin 0 + \cos 0 + C = 1 + C = 1$$
Thus, $C = 0$. The final result is
$$f(x) = \sin x + \cos x$$

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A wheeled inclined plane with an object: Newton's equation of motion

Question:
Consider an inclined cart with motorized wheels as shown. Assume that the coefficients of static and kinetic frictions are equal on the surface of the cart.
(1) After an object is put on the cart gently, it starts falling with acceleration of $a$. The cart is fixed, so it does not move at all. Suppose the inclined angle is the angle where the object just starts falling. Find the coefficient of static friction.
(2) The cart moves ahead and begins accelerated. The, the object becomes stopped falling. Find the acceleration of the cart.

Answer:
(1) From the diagram, we can set up two equations of motion for the moving axis and the axis perpendicular to it.
$$\begin{eqnarray} mg\sin\theta -\mu N &=& ma  \\ N - mg\cos\theta &=& 0 \end{eqnarray}$$
where $\mu$, $N$, and $\mu N$ are the coefficient of static friction, the normal force and the frictional force, $f$. From (2), we obtain the normal force, $N = mg\cos\theta$. Then, equation (1) becomes
$$\begin{equation} mg\sin\theta - \mu mg\cos\theta = ma \end{equation}$$
All $m$'s are cancelled out. Solve for $\mu$.
$$\begin{eqnarray} & & g\sin\theta - \mu g\cos\theta = a  \\ & & \rightarrow \mu g\cos\theta = g\sin\theta - a  \\ & & \rightarrow \mu = \frac{g\sin\theta - a}{g\cos\theta} \end{eqnarray}$$

(2) As shown in the figure, the acceleration of the cart works against the gravitational falling force. The equation of motion in the moving direction becomes
$$\begin{equation} mg\sin\theta - \mu N -ma\cos\theta = 0 \end{equation}$$
The left hand side is the net forces in that direction. The right hand side is zero because there is no motion. Now, set up the other equation of motion for the axis perpendicular to the moving direction.
$$\begin{equation} N - ma\sin\theta - mg\cos\theta = 0 \end{equation}$$
From (8), solve for $N$.
$$\begin{equation} N  = ma\sin\theta + mg\cos\theta \end{equation}$$
Plug this into (7).
$$\begin{eqnarray} & &mg\sin\theta - \mu (ma\sin\theta + mg\cos\theta) -ma\cos\theta = 0  \\ & & \rightarrow g(\sin\theta - \mu\cos\theta) - a(\mu\sin\theta + \cos\theta) = 0  \\ & & \rightarrow a = \frac{\sin\theta - \mu\cos\theta}{\mu\sin\theta + \cos\theta}g \end{eqnarray}$$

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Inscribed and circumscribed circles

Question:
Two circles are given
$$\begin{eqnarray} & & x^2+y^2-6ax+2ay+20a-10 = 0  \\ & & x^2+y^2 = 4 \end{eqnarray}$$
Find $a$ so these circles are inscribed or circumscribed.

Answer:
Equation (1) can be arranged as follows:
$$\begin{eqnarray} & & x^2 -6ax + (3a)^2 -(3a)^2 + y^2 +2ay + (a)^2 -(a)^2 = 10 - 20a  \nonumber \\ & & \rightarrow x^2 -6ax + (3a)^2 + y^2 +2ay + (a)^2  = 10 - 20a + (3a)^2 + (a)^2 \nonumber  \\ & & \rightarrow (x - 3a)^2 + (y + a)^2 = 10a^2 -20a +10  \nonumber \\ & & \rightarrow (x - 3a)^2 + (y + a)^2 = 10(a-1)^2 \end{eqnarray}$$
Therefore, (2) has the center $(0, 0)$ and the radius $2$. Circle (3) has the center $(3a, -a)$ and radius $\sqrt{10}|a-1|$. The distance between those centers is
$$d = \sqrt{(3a-0)^2+(-a-0)^2}=\sqrt{10a^2}=\sqrt{10}|a|$$
If $a=1$, the radius becomes zero. Thus, $a \neq 1$. Let $r_1$ and $r_2$ be radii of two circles. In general, when $d=r_1 + r_2$, they are circumscribed. When $d=|r_1 - r_2|$, they are inscribed. Then, these conditions can be combined as $d = |r_1 \pm r_2|$. Now consider the case in $a>1$. Use the definition above:
$$\begin{equation} \sqrt{10}|a| = | \sqrt{10}(a-1) \pm 2 | \end{equation}$$
Square both sides and solve for $a$.
$$\begin{equation} a = \frac{5 \pm \sqrt{10}}{10} < 1 \end{equation}$$
The result is contradicted. Consider when $a<1$.
$$\begin{equation} \sqrt{10}|a| = | \sqrt{10}(1-a) \pm 2 | \end{equation}$$
Likewise, solve for $a$.
$$\begin{equation} a = \frac{5 \pm \sqrt{10}}{10} \end{equation}$$
which is the proper result.

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Linear transformation: Two dimensional rotation matrix

Question:
There is an ellipse:
$$x^2 + \frac{y^2}{4} = 1$$
If it is on a new coordinate that is rotated by 45$^o$, find the equation of the ellipse with the new coordinate, ($X, Y$).

Answer:
When the coordinate axes are rotated, the old coordinate ($x, y$) will be expressed by the new coordinate ($X, Y$).
$$\left( \begin{array}{c} x \\ y \end{array} \right)  = \left( \begin{array}{cc} \cos\theta & -\sin\theta \\                                       \sin\theta & \cos\theta \end{array} \right) \left( \begin{array}{c} X \\ Y  \end{array} \right)$$
Therefore, we have
$$\left\{ \begin{array}{l} x = X\cos\theta -Y\sin\theta = X\cos 45^o -Y\sin 45^o =\frac{X-Y}{\sqrt{2}} \\ y = X\sin\theta + Y\cos\theta = X\sin 45^o + Y\cos 45^o = \frac{X+Y}{\sqrt{2}} \end{array} \right.$$
Then, they are plugged into the original equation of ellipse.
$$\frac{1}{2}(X-Y)^2 + \frac{1}{8}(X+Y)^2 = 1$$
Or you can also express it as
$$\frac{1}{8}(5X^2 -6XY +5Y^2) = 1$$

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A sliding block on a large cart: Conservation of energy and momentum

Question:
A block B is placed on the cart A that has the frictionless surface and going to be sliding the height of $h$ until the frictional bottom (the shaded part). The cart has frictionless wheels and it moves due to block's falling on the slope. The masses of the block and the cart are $m$ and $M$, respectively. Find each final velocity.

Answer:
The momentum in the horizontal direction should be conserved. Let velocities be $V_A$ and $V_B$. The initial total momentum is zero since they are initially at rest. We assume that the final total momentum is the sum of these momenta. From conservation of momentum, we have
$$\begin{equation} 0 = MV_A+mV_B \end{equation}$$
The total energies in initial and final states must be conserved. The LHS and RHS of the following equation are initial total and final total energies, respectively.
$$\begin{eqnarray}  (0+mgh) + (0 + 0) &=& \left(\frac{1}{2}mV_B^2+0\right) + \left(\frac{1}{2}MV_A^2 + 0 \right) \nonumber  \\ \rightarrow \ \frac{1}{2}MV_A^2 &+& \frac{1}{2}mV_B^2 = mgh \end{eqnarray}$$
Now, derive $V_B$ and $V_A$ from these equations. From (1), we have
$$\begin{equation} V_A = -\frac{m}{M}V_B \end{equation}$$
Plug it in (2).
$$\begin{eqnarray} M \left(\frac{m}{M}\right)^2 V_B^2 + mV_B^2 &=& 2mgh  \\ \left(\frac{m}{M}+1\right)V_B^2 &=& 2gh  \\ V_B = \sqrt{\frac{2Mgh}{m+M}} \end{eqnarray}$$
Then, plug (6) in (3).
$$\begin{equation} V_A = -m\sqrt{\frac{2gh}{M(m+M)}} \end{equation}$$

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Value of an infinite series I

Question:
Consider the following infinite series:
$$S_n = 2-\frac{3}{2}+\frac{3}{2}-\frac{4}{3}+\frac{4}{3}-\frac{5}{4}+ \cdots$$
Find the value(s) when $n \rightarrow \infty$.

Answer:
This series can be generalized as follows:
$$S_n = \sum^{\infty}_{n=1}\left(\frac{n+1}{n}-\frac{n+2}{n+1}\right)$$
The value depends on the number of terms. When $n$ is even, the last term should be $-\frac{n+2}{n+1}$. Namely,
$$\begin{eqnarray} S_n &=& 2-\frac{3}{2}+\frac{3}{2}-\frac{4}{3}+\frac{4}{3}-\frac{5}{4}+ \cdots +\frac{n+1}{n}-\frac{n+2}{n+1}  \\    &=& 2 - \frac{n+2}{n+1} \end{eqnarray}$$
Except the first and last terms, all of them are cennceled out. Therefore, the value of the infinite series is
$$\lim_{n \rightarrow \infty}S_n = \lim_{n \rightarrow \infty}\left(2-\frac{1+\frac{2}{n}}{1+\frac{1}{n}}\right)=1$$
When $n$ is odd, the series ends up with $+\frac{n+1}{n}$. Namely,
$$\begin{eqnarray} S_n &=& 2-\frac{3}{2}+\frac{3}{2}-\frac{4}{3}+\frac{4}{3}- \cdots -\frac{n+1}{n}+\frac{n+1}{n}  \\    &=& 2 \end{eqnarray}$$
Thus we have the value.
$$\lim_{n \rightarrow \infty}S_n = 2$$
The value oscillates between 1 and 2.

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Kinematic diagram: Time vs velocity

Question:
The time vs velocity diagrams of a 10-kg object for $x$ and $y$ directions are given as shown. Describe each motion for each time interval.
(1) 0 < $t$ < 2 seconds
(2) 2 < $t$ < 3 seconds
(3) 3 < $t$ < 4 seconds
(4) at 2 seconds

Answer:
(1) From the diagram, the object moves at the constant velocity, 1 m/s, in $x$ direction, but it gets acceleration of $\frac{3}{2}$ m/s$^2$ in $y$ direction. The slope indicates the acceleration. Thus, it is exerted by 15 N in that direction because $F=ma$.

(2) For both directions, during this time interval, the object moves at the constant velocity. The diagrams show the flat slopes; namely, no acceleration and no force on the object.

(3) The accelerations from 3 s to 4 s are
$$\begin{equation*} a_x = a_y = \frac{0-3}{4-3} = -3 \ \mathrm{m/s^2} \end{equation*}$$
The magnitude of the negative acceleration in both directions is equal, so the force is directed in opposite to 45 degrees on $x-y$ plane. The magnitude of force is given by
$$|F| = 10 \times \sqrt{3^2+3^2} = 42 \ \mathrm{N}$$

(4) At 2 seconds, the velocity in $x$ direction is suddenly changed from 1 m/s to 3 m/s. This indicates the change in momenta; namely, there is an impulse in positive $x$ direction. The impulse is calculated as
$$I = mv_f - mv_i = 10 \cdot 3 - 10 \cdot 1 = 20 \ \mathrm{N\cdot s}$$

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A planet orbits a star in an elliptical orbit: Kepler's second law

Question:
A planet orbits a star in an elliptical orbit. The distance at aphelion is $2a$ and the distance at perihelion is $a$. Find the ratio of the planets speed at perihelion to that at aphelion.

Answer:
According to Kepler's second law, the area swept out per unit time by a radius from the star to a planet is constant. The area can be expressed by
$$dA = \frac{1}{2}r rd\theta$$
Take the derivative in terms of time.
$$\frac{dA}{dt}=\frac{1}{2}r^2 \frac{d\theta}{dt}$$
$\frac{d\theta}{dt}$ is the angular velocity, $\omega$. We know angular momentum $L=mr^2\omega$. Thus,
$$\begin{eqnarray} \frac{dA}{dt} &=& \frac{1}{2}r^2\omega  \\   &=& \frac{L}{2m} \end{eqnarray}$$
This shows that the equal area per unit time indicates constant angular momentum, $L$. We get back to the original expression of the angular momentum, $L=mr^2\omega$. We also know that $v=r\omega$ and $v$ is the tangential speed, so $L=mvr$. Compare the angular momenta at perihelion and aphelion.
$$L = mv_{\mathrm{p}}a = mv_{\mathrm{a}}2a$$
Therefore the ratio of the speeds is
$$v_{\mathrm{p}}:v_{\mathrm{a}}=2:1$$
We can state that Kepler's second law is essentially equal to the conservation of angular momentum. The more radius, the slower the planet has. The less radius, the faster the planet gets.

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Gravitational field at height from the surface of a planet

Question:
Find the magnitude of the gravitational acceleration near the surface of a planet of radius $R$ at height, $h$ to the second order. Let $g_0$ be the gravitational acceleration at $h=0$.

Answer:
Use the universal law of gravitation.
$$mg = \frac{GMm}{r^2}$$
So
$$g = \frac{GM}{r^2}$$
The distance $r$ is the radius of the planet and the height, $r=R+h$
$$g = \frac{GM}{(R+h)^2}$$
We can arrange it as follows:
$$\begin{eqnarray} & & g  = \frac{GM}{R^2}\frac{1}{\left(1+\frac{h}{R} \right)^2}  \\ & & g = g_0 \frac{1}{\left(1+\frac{h}{R} \right)^2} \end{eqnarray}$$
Since $h \ll R$, we can use expansion, $\sum^{\infty}_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^n$. The second order of the approximation is
$$g=g_0 \left[ 1 - 2\frac{h}{R} + 3\left(\frac{h}{R}\right)^2 \right]$$

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Conservation of energy and spring and pendulum bob

Question:
A pendulum bob is released from the height of $h$ to hit a spring that creates the force $F=-kx-bx^3$ in terms of the displacement. If the pendulum has mass $m$, find the compression displacement of the spring.

Answer:
Find the potential energy of spring. Since it is a conservative force, we integrate it in terms of displacement.
$$U = -\int F dx = \int kx+bx^3 dx = \frac{1}{2}kx^2+\frac{1}{4}bx^4$$
The potential energy of the bob is $mgh$. This can be transferred into the spring energy, so
$$mgh =  \frac{1}{2}kx^2+\frac{1}{4}bx^4$$
Rearrange it to solve for $x$:
$$\begin{eqnarray} & & x^4 +\frac{2k}{b}x^2 = \frac{4mgh}{b}  \\ & & \left(x^2+\frac{k}{b}\right)^2 - \frac{k^2}{b^2} = \frac{4mgh}{b}  \\ & & x^2 + \frac{k}{b} = \sqrt{\frac{4mgh}{b}+\frac{k^2}{b^2}}  \\ & & x  = \sqrt{\sqrt{\frac{4mgh}{b}+\frac{k^2}{b^2}} -\frac{k}{b}} \end{eqnarray}$$

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Equation for a linear line: Knowing the coordinates of two points

Question:
A line goes through points (-3, 9) and (2, 4). Find the equation of the linear line.

Answer:
There are two methods. You can use a line equation with two unknowns:
$$y = ax + b$$
where $a$ is the slope and $b$ is the intercept. Plug in the coordinates to solve for simultaneous equations.
$$\begin{eqnarray} 9 &=& -3a + b \\ 4 &=& 2a + b \end{eqnarray}$$
(1)-(2) gives $5 = -5a$. Thus, $a=-1$. Plug back in either equation and we get $b=6$. The line equation is
$$y = -x + 6$$
The other method is to utilize following formula:
$$(y-y_1)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$$
The factor, $\frac{y_2-y_1}{x_2-x_1}$, corresponds to the slope. After plugging in the coordinates, we can obtain the same result.

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Gravitational force: Moon causing a tidal force on the Earth's ocean

Question:
The gravitational force between the moon and Earth creates a tidal force. From the figure, $a$ is the distance between the moon and the Earth. $M$ and $m$ are the masses of Earth and moon, respectively. $r$ denotes the radius of Earth. Find the differential tidal acceleration.

Answer:
The tidal force is obtained by the difference of gravitational fields between C (center of mass) and S (place to get tidal force). This can be associated with the differential tidal acceleration. Let us write down each gravitational acceleration.
$$\begin{eqnarray} g_C &=& \frac{Gm}{a^2}  \\ g_S &=& \frac{Gm}{(a+r)^2} \end{eqnarray}$$
The difference of them is the tidal acceleration.
$$\begin{eqnarray} g_C -g_S &=& \frac{Gm}{a^2} - \frac{Gm}{(a+r)^2}  \\    &=& \frac{Gm}{a^2}\left(1-\frac{a^2}{(a+r)^2}\right)  \\    &=& \frac{Gm}{a^2}\left(1-\frac{1}{(1+\frac{r}{a})^2}\right)  \\    &\sim& \frac{Gm}{a^2}\left(1-\left\{1-2\frac{r}{a}\right\}\right) \end{eqnarray}$$
The above uses approximation. Hence, we have
$$g_{\mathrm{tidal}} = \frac{2Gmr}{a^3}$$

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Gravitational field of a hollow sphere

Question:
A hollow sphere has region $a<r<b$ filled with mass of uniform density $\rho$. Find the magnitude of the gravitational field between $a$ and $b$.

Answer:
Utilize Gauss's  law for gravitational fields.
$$\int_{S} g d\alpha = -4\pi GM$$
As we know, if there is no mass in a sphere, no gravitational field is detected. Thus, when $r<a$, $g=0$. We can also find the field when $r>b$.
$$\int_{S} g d\alpha = -4\pi GM  \\ \rightarrow 4 \pi r^2 g = -4 \pi GM  \\ \rightarrow g = \frac{GM}{r^2}$$
The integral of left hand side gives surface area of a sphere. For $a<r<b$, the mass, $M$, depends on the volume.
$$M_{a-b} = \int \rho dV = \rho \int^{r}_{a}r^2 \int^{\pi}_{0}\sin \theta d\theta \int^{2\pi}_{0}d\phi \\ =\rho\frac{4\pi}{3}(r^3-a^3)$$
From Gauss's law,
$$-4\pi r^2 g = -4\pi G \rho\frac{4\pi}{3}(r^3-a^3)$$
Therefore, we have the gravitational field in $a<r<b$.
$$g = \frac{4\pi}{3}G\rho\left(r-\frac{a^3}{r^2}\right)$$

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Radio active decay: Half life

Question: 
Find the half life for the process of radioactive decay. Assume that $\lambda$ is the decay constant.

Answer:
The radioactive decay is understood as the comparison of the numbers of nucleus at time $t$ and time $t+\Delta t$. The difference of the numbers can be expressed as $N(t)-N(t+\Delta t)$, and this is proportional to a rate, $\lambda$.
$$N(t)-N(t+\Delta t) = -\lambda N(t) \Delta t$$
The right hand side indicates how the number decreases for $\Delta t$. We can rearrange it as
$$\frac{N(t)-N(t+\Delta t)}{\Delta t}=\frac{dN}{dt}=-\lambda N(t)$$
This is a first order of differential equation with respect to time. Again, rearrange and integrate it as follows:
$$\int^N_{N_0} \frac{dN}{N}=\int^{t}_{0} -\lambda dt$$
This gives
$$\ln \frac{N}{N_0} = -\lambda t \quad \rightarrow \quad N=N_0e^{-\lambda t}$$
In order to find the half life, let $N=\frac{N_0}{2}$. Then derive the time, $t_{0.5}$.
$$\begin{eqnarray*} & & \frac{N_0}{2}=N_0e^{-\lambda t_{0.5}}  \\ \rightarrow & & \frac{1}{2} = e^{-\lambda t_{0.5}}  \\ \rightarrow & & 2 = e^{\lambda t_{0.5}}  \\ \rightarrow & & \lambda t_{0.5} = \ln 2  \\ \rightarrow & & t_{0.5} = \frac{\ln 2}{\lambda} \end{eqnarray*}$$
Thus, the half life is given by $\frac{0.693}{\lambda}$.

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A binary star system and Kepler's law

Question:
Consider a binary star system to find the mass of the stars. The distance between them is found out to be $a$, and the period of revolution $T$. Assume that the masses of two stars are equal. Find the mass from the conditions.

Answer:
This method allows us to find the total mass of the system in laboratory by knowing the distance between stars and the period. Kepler's third law indicates the relationship between the period and distance.
$$\frac{T^2}{a^3} = \frac{4\pi^2}{G(m_1+m_2)}$$
where $G$ is the gravitational constant. Since $m_1+m_2 = 2m$, we can solve for the mass.
$$m = \frac{2\pi^2 a^3}{GT^2}$$
This gives the mass of one star.

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Inequality of trigonometric functions

Question:
(1) The range of $x$ is: $0^o \leqq x \leqq 180^o$. Solve $\cos^4 x > \sin^4 x$.
(2) The range of $x$ is: $0^o \leqq x < 360^o$. Solve $\cos 2x + \sin x < 0$.

Answer:
(1) Arrange the equation.
$$\begin{eqnarray} \cos^4 x - \sin^4 x &>& 0  \\ (\cos^2 x + \sin^2 x)(\cos^2 x - \sin^2 x) &>& 0  \\ (1)(\cos x + \sin x)(\cos x - \sin x) &>& 0 \end{eqnarray}$$
Therefore, we can state
$$\begin{eqnarray} & & (\cos x + \sin x)>0 \ \mathrm{and} \ (\cos x - \sin x)>0  \\ \mathrm{or}  & & \nonumber \\ & & (\cos x + \sin x)<0 \ \mathrm{and} \ (\cos x - \sin x)<0 \end{eqnarray}$$
In case of (4), within the $x$ range, when $\cos x > \sin x$, $0^o \leqq x < 45^o$. When $\cos x + \sin x > 0$, $0^o \leqq x < 135^o$ since the solution of $\cos x + \sin x = 0$ is $x=135^o$. Therefore, one range is
$$0^o \leqq x < 45^o$$
In case of (5), when $\cos x < \sin x$, $45^o < x \leqq 180^o$. When $\cos x + \sin x < 0$, $135^o < x \leqq 180^o$. Therefore, the other range is
$$135^o < x \leqq 180^o$$
Thus, the answer is $0^o \leqq x < 45^o$ and $135^o < x \leqq 180^o$.

(2) Use the double angle formula, $\cos 2x = 1-2\sin^2 x$ to rearrange the given expression.
$$\begin{eqnarray} 1-2\sin^2 x +\sin x &<& 0 \\ 2\sin^2 x -\sin x-1 &>& 0 \\ (2\sin x + 1)(\sin x -1) &>& 0 \end{eqnarray}$$
$\sin x -1$ is always negative or zero, so the condition must be only
$$2\sin x + 1 <0 \ \mathrm{and} \ \sin x -1<0$$
Then, $2\sin x + 1 <0$ determines the range. Namely, $\sin x < -\frac{1}{2}$. For $\sin x = -\frac{1}{2}$, $x= -30^o$ or $210^o$. Therefore, the range must be
$$210^o < x < 330^o$$

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Rigid body: A circular cylinder on an inclined plane

Question:
A circular cylinder of radius $r$ rolls down on an inclined plane from height $h$. Compare the velocities at the bottom of this object with that of a point object.

Answer:
The moment of inertia of a cylinder (disk) is given by
$$I = \frac{1}{2}mr^2$$
The derivation is provided on this webpage: http://hirophysics.com/Study/moment-of-inertia.pdf Use conservation energy. For this cylinder, we need to include the rotational kinetic energy in addition to linear kinetic energy. We suppose that the entire potential energy is transferred into all the kinetic energy.
$$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$$
Since $\omega=\frac{v}{r}$ and $I = \frac{1}{2}mr^2$, we substitute it in the above, then solve for the terminal velocity.
$$\begin{eqnarray*} & &\frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2 = mgh  \\ & & \frac{3}{4}mv^2=mgh \\ & & v_C = \sqrt{\frac{4gh}{3}} \end{eqnarray*}$$
$v_C$ is the final velocity of the cylinder. Let us find the final velocity of a point object. Likewise, we use conservation of energy, but the rotational kinetic energy is excluded.
$$\begin{eqnarray*} & &\frac{1}{2}mv^2 = mgh  \\ & & v_P = \sqrt{2gh} \end{eqnarray*}$$
Now compare them.
$$\frac{v_C}{v_P}=\frac{\sqrt{\frac{4gh}{3}}}{\sqrt{2gh}}=\sqrt{\frac{2}{3}}\sim 0.816$$
The final velocity of the cylinder is slower than that of point object by the factor of 0.816.

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Magnetic field in the hydrogen atom with Bohr theory

Question:
Use Bohr theory. Find the magnetic field by electron's exerting at the nucleus of the hydrogen atom. The radius of hydrogen is assumed to be 0.529 $\times$ 10$^{-10}$ m.

Answer:
We can assume that the electron moves around the nucleus as making a circular current. According to the Biot-Savart Law, we have the magnetic field of a circular current as
$$B_z = \frac{\mu_0 I}{4 \pi}\int^{2\pi r}_{0}\frac{dl}{r^2}=\frac{\mu_0 I}{2r}$$
Now, we need to obtain the current. The electric current is defined by $I=\frac{Q}{T}$. The charge of an electron or a proton is 1.61 $\times$ 10$^{-19}$ C. In order to get the current, we have to find the time period. That is
$$T=\frac{2\pi r}{v}$$
where $v$ is the tangential velocity of the electron. So the current is expressed by
$$I = \frac{Q}{\frac{2\pi r}{v}}$$
In terms of the classical theory, the centripetal force is equal to the Coulomb force. Then, solve for the velocity.
$$\begin{eqnarray} \frac{mv^2}{r}&=&\frac{ke^2}{r^2} \\ v&=&\sqrt{\frac{ke^2}{mr}}  \\   &=& \sqrt{\frac{8.99 \times 10^9 (1.61 \times 10^{-19})^2}{9.11 \times 10^{-31}\cdot 0.529 \times 10^{-10}}}  \\   &=& 2.20 \times 10^6 \ \mathrm{m/s} \end{eqnarray}$$
Therefore, we calculate the current.
$$I=\frac{Q}{\frac{2\pi r}{v}}=\frac{1.61 \times 10^{-19}}{\frac{2\pi \times 0.529\times 10^{-10}}{2.20\times 10^6}}=1.07\times 10^{-3} \ \mathrm{A}$$
Then, we can find the magnetic field.
$$B=\frac{\mu_0 I}{2r}=\frac{4\pi \times 10^{-7} \times1.07\times 10^{-3}}{2 \times 0.529\times 10^{-10}}= 12.7 \ \mathrm{T}$$

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Motion with mass decreasing: A rocket launch

Question:
A rocket is launched in a constant gravity, 9.80 m/s$^2$. The initial velocity is 400 m/s, and the burn time is 100 seconds. If the exhaust velocity is 2000 m/s, and the mass decreases by a factor of three; namely, the current mass divided by the original mass is equal to $\frac{1}{3}$, find the final velocity of the rocket.

Answer:
The equation of motion is given as
$$\begin{equation} m\frac{dv}{dt}=-mg-u \frac{dm}{dt} \end{equation}$$
where $u$ is the exhaust velocity. Divide both sides
by the mass, $m$.
$$\begin{equation} \frac{dv}{dt}=-g-u \frac{dm}{dt}\frac{1}{m} \end{equation}$$
Multiply $dt$ by both sides. (This manipulation is not for mathematicians.)
$$\begin{equation} dv=-gdt - u \frac{dm}{m} \end{equation}$$
Integrate this.
$$\begin{equation} \int^{v}_{v_0} dv=\int^{t}_0 -gdt - u \int^{m}_{m_0}\frac{dm}{m} \end{equation}$$
Therefore,
$$\begin{equation} v-v_0 = -gt - u \ln \frac{m}{m_0} \end{equation}$$
Plug in the numbers.
$$\begin{equation} v = 400 - 9.80 \cdot 100 - 2000 \ln \frac{1}{3}=1620 \ \mathrm{m/s} \end{equation}$$

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Projectile motion: With air resistance proportional to the velocity

Question:
A ball is thrown and it tracks a projectile near the Earth's surface subject to a resistive force due to air, $f_R = -bv$, which is proportional to its velocity. Find the velocities, positions for $x$ and $y$ directions; and the time of flight.

Answer:
Newton's second law shows
$$\begin{eqnarray} mv'_x &=& -bv_x  \\ mv'_y &=& -mg-bv_y \end{eqnarray}$$
Solve the equation of motion in $x$ direction.
$$\begin{equation} v_x = v_{0x}e^{-\frac{b}{m}t} \quad \rightarrow \quad v_x = v_{0x}e^{-\gamma t} \end{equation}$$
where $\gamma=\frac{b}{m}$.
Integrate it again with respect to time.
$$\begin{eqnarray} x &=& v_{0x}\int e^{-\gamma t} dt  \\    &=& v_{0x} \left[ -\frac{1}{\gamma}e^{-\gamma t}\right]^t_0  \\   &=& \frac{v_{0x}}{\gamma}(1-e^{-\gamma t}) \end{eqnarray}$$
Likewise, we can integrate the $y$ component of the equation.
$$\begin{equation} \frac{dv_y}{dt} = -g  - \gamma v_y \end{equation}$$
Then, we separate the terms for $v$ and the other.
$$\begin{equation} \frac{dv_y}{dt} = - \gamma \left(v_y +\frac{g}{\gamma}\right) \end{equation}$$
Now, divide it by $v_y +\frac{g}{\gamma}$ and multiply by $dt$ although this statement is not recommended for talking to mathematicians.
$$\begin{eqnarray} & & \frac{dv}{v_y + \frac{g}{\gamma}} = -\gamma dt  \\ & & \ln\left| v_y +\frac{g}{\gamma}\right| = -\gamma t + C \\ & & v_y + \frac{g}{\gamma} = e^{-\gamma t + C}\\ & & v_y = Ce^{-\gamma t} - \frac{g}{\gamma} \end{eqnarray}$$
Consider the initial condition: When $t=0$, $v_y = v_{0y}$. Therefore, $C=v_{0y}+\frac{g}{\gamma}$. We now have
$$\begin{equation} v_y = \left(v_{0y}+\frac{g}{\gamma}\right)e^{-\gamma t} - \frac{g}{\gamma} \end{equation}$$
Integrate it again with time.
$$\begin{eqnarray} y &=& \int \left\{\left(v_{0y}+\frac{g}{\gamma}\right)e^{-\gamma t} - \frac{g}{\gamma}\right\} dt \\   &=& \left[ -\frac{1}{\gamma}\left(v_{0y}+\frac{g}{\gamma}\right)e^{-\gamma t} - \frac{gt}{\gamma}\right]^t_0  \\   &=& \frac{1}{\gamma}\left(v_{0y}+\frac{g}{\gamma}\right)(1-e^{-\gamma t}) - \frac{gt}{\gamma} \end{eqnarray}$$
When $y=0$, we can obtain the entire time of flight. Thus,
$$\begin{eqnarray} \frac{gt}{\gamma} &=& \frac{1}{\gamma}\left(v_{0y}+\frac{g}{\gamma}\right)(1-e^{-\gamma t}) \\ \frac{g}{\gamma} &=& \left(v_{0y}+\frac{g}{\gamma}\right)\frac{1-e^{-\gamma t}}{\gamma t} \\ \frac{1-e^{-\gamma t}}{\gamma t} &=& \frac{g}{\gamma\left( v_{0y}+ \frac{g}{\gamma}\right)} \\ \frac{1-e^{-\gamma t}}{\gamma t} &=& \frac{1}{1+ \frac{\gamma v_{0y}}{g}} \end{eqnarray}$$
Expand both sides by assuming that $\gamma \ll 1$.
$$\begin{equation} {\textstyle 1-\frac{1}{2!}\gamma t+\frac{1}{3!}(\gamma t)^2-\frac{1}{4!}(\gamma t)^3+\cdots = 1-\frac{\gamma v_{0y}}{g}+\left(\frac{\gamma v_{0y}}{g}\right)^2-\left(\frac{\gamma v_{0y}}{g}\right)^3 + \cdots} \end{equation}$$
Let $G=\frac{\gamma v_{0y}}{g}$, and expand $\gamma t$ in terms of polynomial of $G$. Namely,
$$\begin{equation} \gamma t=a_1 G+a_2 G^2 + a_3 G^3 \cdots \end{equation}$$
Plug this in the left hand side of (21). Let us take it up to $a_2$ and compare them with the right hand side in terms of $G^n$. Then, we have $a_1 = 2$ and $a_2 = -\frac{2}{3}$. Again, plug this in (22).
$$\begin{equation} \gamma t = \frac{2v_{0y} \gamma}{g}-\frac{2}{3}\frac{\gamma^2 v_{0y}^2}{g^2}+\cdots \end{equation}$$
Therefore the time $t=T$ is approximately given as follows:
$$T = \frac{2v_{0y}}{g}-\frac{2}{3}\frac{\gamma v_{0y}^2}{g^2}$$

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Natural frequency: Two objects with a spring

Question:
Two objects which have equal masses, $m$, are connected with a spring whose force constant is $k$. The objects are placed on a frictionless surface and oscillating without external forces. Find the natural frequency of this motion.

Answer:
$x_1$ and $x_2$ are the displacements of each object from the natural length. The displacement, $x$, is the one for the spring, so $x=x_1-x_2$ for object 1 and $-x=x_2-x_1$ for object 2. Set up the equations of motion:
$$\begin{eqnarray} mx_1'' &=& -kx  \\ mx_2'' &=& kx \end{eqnarray}$$
Subtract (2) from (1).
$$\begin{equation} m(x_1''-x_2'') = -2kx \end{equation}$$
Since $x'' = x_1''-x_2''$, we have
$$\begin{equation} mx'' + 2kx = 0 \quad \rightarrow \quad x'' + \frac{2k}{m} = 0 \end{equation}$$
The solution of this differential equation is
$$\begin{equation} x = A\sin\left(\sqrt{\frac{2k}{m}} t\right) + B\cos\left(\sqrt{\frac{2k}{m}} t\right) \end{equation}$$
where $\sqrt{\frac{2k}{m}}$ is the natural frequency of this spring motion.

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Uncertainty principle: Time and energy with Delta baryon

Question:
The Delta baryon, $\Delta$, is known as the lowest nucleon resonance that has a mass of 1232 MeV/c$^2$ and a width of 120 MeV/c$^2$. Its spin and isospin are equally 3/2. Find the lifetime of this particle.

Answer:
According to the uncertainty principle between time and energy, we have
$$\Delta E \cdot \Delta t \geqq \frac{\hbar}{2}$$
The multiplication of uncertainties is greater than or equal to a half of the reduced Planck constant (or Dirac's constant). Solve for the time and plug in numbers. In this unit system, the constant $\hbar$ should be $\hbar \times c$ = 197.33 MeV fm, where fm is 10$^{-15}$ m. Therefore,
$$\Delta t \sim \frac{\hbar c}{2\Delta E c} = \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{MeV \ fm}{MeV / c}} = \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{fm}{c}}  \\ = \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{10^{-15}m}{3.00 \times 10^8 m/s}} \\ = 2.74 \times 10^{-24} \ \mathrm{s}$$

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Special theory of relativity: Time dilation in an airplane

Question:
Physicists conduct an experiment to measure time dilation of an atomic clock in a flying airplane. If the velocity of the airplane is 960 km/h, find the time dilated of the atomic clock from the laboratory flame. (You do not need to consider the effect from the general theory of relativity.)

Answer:
We consider one direction of motion, say, $x$ direction. The Lorentz transformations for the displacement and time are
$$\begin{eqnarray} x' &=& \frac{-vt+x}{\sqrt{1-\beta^2}}  \\ t' &=& \frac{t-(v/c^2)x}{\sqrt{1-\beta^2}} \end{eqnarray}$$
where $\beta=v^2/c^2$. Suppose that the clock is placed at the origin of moving flame, $S'$. Then, one measures the time in laboratory flame, $S$. The origin in $S'$ indicates that $x'=0$. Thus, we have
$$\begin{equation} 0 = \frac{-vt+x}{\sqrt{1-\beta^2}} \quad \rightarrow \quad x = vt \end{equation}$$
Plug it in the transformation for time.
$$\begin{equation} t' = \frac{t-(v/c^2)(vt)}{\sqrt{1-\beta^2}}=\frac{t(1-\beta^2)}{\sqrt{1-\beta^2}}=\sqrt{1-\beta^2}t \end{equation}$$
Since $v\ll c \rightarrow \beta \ll 1$, we can use the approximated expression with a Taylor expansion;
$$t' = \left(1-\frac{1}{2}\beta^2\right)t$$
Now, convert the velocity of the airplane into m/s. Note that 1 km = 1000 m and 1 hour = 3600 s.
$$960 \ \mathrm{km/h} = 960 \times 1000 \div 3600 = 266.7 \ \mathrm{m/s}$$
Therefore, we have the time dilation compared with the lab frame.
$$t'= \left\{1-\frac{1}{2}\left(\frac{266.7}{3.00\times 10^8}\right)^2\right\}t=(1-3.95\times 10^{-13})t$$
This means: While one second elapses in lab frame, the clock in the airplane only elapses $1-3.95\times 10^{-13}$ seconds.

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A special type of differential equation III

Question:
Solve $(x^2\cos x - y)dx + xdy = 0$.

Answer:
Rewrite the equation.
$$\frac{dy}{dx}-\frac{y}{x}=-x\cos x$$
When the right hand side is zero, the solution is $y=Cx$. Using variation of constants, we make $C$ as a function of $x$. Namely, $y=C(x)x$ and plug it in the above equation.
$$\begin{eqnarray} & & \frac{d(C(x)x)}{dx}-\frac{C(x)x}{x}=-x\cos x  \\ & & C'(x)x+C(x)-C(x)=-x\cos x  \\ & & C'(x) = - \cos x  \\ & & C(x) = - \sin x + C \end{eqnarray}$$
We know $C(x)=\frac{y}{x}$, so
$$\begin{eqnarray} & & \frac{y}{x} = - \sin x + C \\ & & y = x(-\sin x + C) \end{eqnarray}$$

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A special type of differential equation II

Question:
(1) Solve
$x\frac{dy}{dx}+y=0$
(2) Solve
$x\frac{dy}{dx}+y=x\ln x$ by using the result from (1).

Answer:
(1) The differential equation can be expressed as
$$x\frac{dy}{dx}+\frac{dx}{dx}y =0, \qquad \rightarrow  x'y + xy' =0$$
Namely,
$$\frac{d(xy)}{dx}=0 \qquad \rightarrow xy=C \quad \rightarrow \rightarrow y=\frac{C}{x}$$

(2) Use variation of constants. When $x \ln x = 0$, $y=\frac{C(x)}{x}$. Then, plug it in the original equation.
$$\begin{eqnarray*} & & x\frac{d}{dx}\frac{C(x)}{x}+\frac{C(x)}{x}=x \ln x  \\ & & \frac{xC'(x)-C(x)}{x^2}x+\frac{C(x)}{x}=x\ln x  \\ & & C'(x) = x \ln x \end{eqnarray*}$$
Integrate both sides.
$$\begin{eqnarray*} C(x) &=& \int x\ln x dx  \\ C(x) &=& \frac{x^2}{2}\ln x - \int \left(\frac{1}{x} \frac{x^2}{2}\right)dx  \\ C(x) &=&  \frac{x^2}{2}\ln x - \frac{x^2}{4}+ C \end{eqnarray*}$$
Since $C=xy$, we have
$$\begin{eqnarray*} xy &=&  \frac{x^2}{2}\ln x - \frac{x^2}{4} + C \\ y &=& \frac{x}{2}\ln x - \frac{x}{4} + \frac{C}{x} \end{eqnarray*}$$

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A special type of differential equation I

Question:
(1) Find the general solution of
$\frac{dy}{dx}+P(x)y=Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$.
(2) Solve the following differential equation:
$x\frac{dy}{dx}+2y=\sin x$

Answer:
(1) Multiply $e^{\int P(x)dx}$ by both sides.
$$e^{\int P(x)dx}\frac{dy}{dx}+P(x)e^{\int P(x)dx}y=e^{\int P(x)dx}Q(x)$$
We can notice that the left hand side can be modified as follows:
$$\frac{d}{dx}\left(e^{\int P(x)dx}y\right) = e^{\int P(x)dx}Q(x)$$
Integrate both sides in terms of $x$.
$$\begin{eqnarray*} e^{\int P(x)dx}y &=& \int e^{\int P(x)dx}Q(x)dx + C \\ y &=& e^{-\int P(x)dx}\left(\int e^{\int P(x)dx}Q(x)dx + C \right) \end{eqnarray*}$$

The other solution of (1) When $Q(x) = 0$, the solution is $y=Ce^{-\int P(x)dx}$. Therefore, when $Q(x) \neq 0$, the constant $C$ can be a function of $x$ and then find the $C(x)$. $y=Ce^{-\int P(x)dx}$ can be plugged in the original equation:
$$\begin{eqnarray*} \frac{d}{dx}Ce^{-\int P(x)dx}+P(x)y&=&Q(x)  \\ C'e^{-\int P(x)dx}-CP(x)e^{-\int P(x)dx}+P(x)y&=&Q(x) \\ C'e^{-\int P(x)dx}-P(x)y+P(x)y&=&Q(x) \\ C'&=&Q(x)e^{\int P(x)dx}  \\ C(x) &=& \int Q(x)e^{\int P(x)dx}dx+C \end{eqnarray*}$$
Since $C(x)=e^{\int P(x)dx}y$, we have
$$\begin{eqnarray*} e^{\int P(x)dx}y &=& \int Q(x)e^{\int P(x)dx}dx+C \\ y &=& e^{-\int P(x)dx}\left(\int Q(x)e^{\int P(x)dx}dx+C \right) \end{eqnarray*}$$

(2) The given equation can be modified as
$$\frac{dy}{dx}+\frac{2y}{x}=\frac{\sin x}{x}$$
From the previous discussion, $P(x)=2/x$ and $Q(x)=\sin x/x$. So plug them in to the previous result.
$$y = e^{-\int \frac{2}{x}dx}\left(\int \frac{\sin x}{x}e^{\int \frac{2}{x}dx}dx+C \right)$$
Think about the function $f(x)=e^{-\int \frac{2}{x}dx}$. This becomes $f(x)=e^{-2\ln x}$. In order to simplify it, take log of both sides.
$$\begin{eqnarray*} \ln f(x) &=& \ln e^{-2\ln x}  \\ \ln f(x) &=& -2\ln x  \\ \ln f(x) &=& \ln x^{-2}  \\ f(x) &=& x^{-2} \end{eqnarray*}$$
Thus, we have
$$y = \frac{1}{x^2}\left(\int \frac{\sin x}{x}x^2 dx+C \right)$$
For the final steps,
$$\begin{eqnarray} y &=& \frac{1}{x^2}\left(\int x\sin x dx+C \right)  \\ &=&  \frac{1}{x^2}\left([-x\cos x + \int\cos x dx]+C \right)  \\ &=&  \frac{1}{x^2}\left(-x\cos x + \sin x + C \right) \end{eqnarray}$$

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A rope unwounded from a cylindrical wheel

Question:
A cylindrical wheel with radius of 2-m rotates on a fixed frictionless horizontal axis. The moment of inertia of the wheel is 10.0 kg m$^2$. A constant tension on the rope exerted around the rim is 40.0 N. If the wheel starts from rest at $t=0$ s, calculate the length of the rope unwounded after 3.00 seconds.

Answer:
The equation of motion for rotation is
$$\sum \tau = rF = I\alpha$$
where $\tau$ is the torque, $F$ is the tension, $I$ is the moment of inertia, and $\alpha$ is the angular acceleration. Now, solve for the angular acceleration.
$$\begin{equation} \alpha = \frac{rT}{I} = \frac{2\cdot 40}{10} = 8.00 \ \mathrm{rad/s^2} \end{equation}$$
According to angular kinematics, we can calculate the angular displacement.
$$\begin{equation} \theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2 = 0+0+\frac{1}{2}\alpha t^2 =\frac{1}{2}\cdot 8.00 \cdot 3^2 = 36.0 \ \mathrm{rad} \end{equation}$$
The linear displacement is
$$d = r\theta = 2.0 \cdot 36.0 = 72.0 \ \mathrm{m}$$

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Resistor and inductor circuit with a steady voltage source

Question:
A circuit that has a resistor, $R$, and an inductor, $L$, which are connected in series with a DC voltage source $V$. The initial value of the current is $I(t=0)=0$ A. Find the current as a function of time.

Answer:
From Kirchoff's law, the generated voltage is consumed by each element.
$$V-RI-L\frac{dI}{dt}=0$$
$$\begin{eqnarray} RI+L\frac{dI}{dt} &=& V  \\ I' + \frac{1}{\tau}I &=& \frac{V}{L} \end{eqnarray}$$
where $I'=\frac{dI}{dt}$ and $\tau=\frac{L}{R}$. Now, let $D\equiv \frac{d}{dt}$.
$$\begin{equation} \left(D+\frac{1}{\tau}\right)I=\frac{V}{L}  \\ \end{equation}$$
The fundamental solution is
$$\begin{equation} I_f = C_1e^{-\frac{t}{\tau}} \end{equation}$$
The particular solution is
$$\begin{equation} I_p = \frac{1}{D+\frac{1}{\tau}}\cdot \frac{V}{L}=\frac{1}{0+\frac{1}{\tau}}\cdot \frac{V}{L}=\frac{V}{R} \end{equation}$$
Therefore, we have the general solution.
$$\begin{equation} I = C_1e^{-\frac{t}{\tau}} + \frac{V}{R} \end{equation}$$
When $t=0$, $I=0$. So $0=C_1+V/R$ and $C_1=-V/R$. Then, the current is expressed as
$$I(t) = -\frac{V}{R}e^{-\frac{t}{\tau}}+\frac{V}{R}$$
or
$$I(t) = \frac{V}{R}(1-e^{-\frac{t}{\tau}})$$

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Ampere's law: A toroid carrying current

Question:
A toroid has $N$ turns both sides carrying equal current $I$. The inner and outer radii are $a$ and $b$, respectively. A distance $r$ is supposed to be exact the middle between $a$ and $b$. Find the magnetic field at $r$.

Answer:
From Ampere's law, if inside the contour has a current $I$, we have the relationship:
$$\oint \vec{B}\cdot dl = \mu_0 I$$
The radius of the contour is $r$ which encloses the current $N\times I$. Therefore,
$$B(2\pi r) = \mu_0 NI$$
Recall that $r=\frac{a+b}{2}$; thus, the magnetic field is
$$B = \frac{\mu_0 NI}{\pi(a+b)}$$

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Finding the enclosed area: A parabola and a circle

Question:
A parabola is given as
$$\begin{equation} y=\frac{1}{4}x^2-1 \end{equation}$$
A circle is also given in the same plane:
$$\begin{equation} x^2+y^2=16 \end{equation}$$
Calculate the enclosed area of the upper part segmented by the parabola and the circle.

Answer:
Find the intersections first
. From equation (2), the ranges of $x$ and $y$ are $-4\leqq x,y \leqq 4$. Solve for $x^2$ in (2).
$$\begin{equation} x^2=16-y^2 \end{equation}$$
Plug it in (1)
$$\begin{eqnarray} & & y = \frac{1}{4}(16-y^2)-1  \\ & & y^2 = 4y -12 = 0  \\ & & (y-2)(y+6) = 0 \end{eqnarray}$$
Therefore, $y=2$ or $y=-6$; however, $y=-6$ is beyond the range. $y$ takes only 2 for the coordinate. Using this result, we obtain $x$ as $\pm 2\sqrt{3}$, and both values are within the range. The coordinates of the intersections are ($+2\sqrt{3}, 2$) and ($-2\sqrt{3}, 2$). Now, find the area of the shaded part. As shown in the figure, it is subtracted the right angle triangle from the sector with 60 degrees. In other expression, (Area of the shaded part) = (Area of the sector) - (Area of the triangle). That is
$$\begin{eqnarray} &=& \left(\pi 4^2 \times \frac{60}{360}\right) - \left(\frac{1}{2}\times 2 \times 2\sqrt{3}\right)  \\ &=& \frac{8}{3}\pi - 2\sqrt{3} \end{eqnarray}$$
The lower part is constructed by the functions between $y=2$ and $y=\frac{1}{4}x^2-1$.
$$\begin{eqnarray} & &\int^{2\sqrt{3}}_0 \left\{2-\left(\frac{1}{4}x^2-1\right)\right\}dx  \\ & & = \left[3x-\frac{1}{4}\cdot \frac{x^3}{3}\right]^{2\sqrt{3}}_0  \\ & & = 3 \times 2\sqrt{3} - \frac{1}{4} \times \frac{(2\sqrt{3})^3}{3}  \\ & & = 4\sqrt{3} \end{eqnarray}$$
Then, the upper shaded and lower parts are added.
$$A_{\mathrm{half}}=\frac{8}{3}\pi - 2\sqrt{3}+4\sqrt{3}=\frac{8}{3}\pi + 2\sqrt{3}$$
Therefore, the area is
$$A = 2 \times A_{\mathrm{half}}=\frac{16}{3}\pi + 4\sqrt{3}$$

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Special theory of relativity: Velocity addition

Question:
The speed of light is measured in a liquid that has refractive index of $n$. If the liquid is moving at $v$, what is the speed of light in the liquid detected in the laboratory frame?

Answer:
The addition of two velocities from the lab frame in terms of relativity is
$$v_{\mathrm{added}} = \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$$
The speed of light in liquid is given by
$$v' = \frac{c}{n}$$
In this case, $v_1=\frac{c}{n}$ and $v_2=v$.
$$\begin{eqnarray} v_{\mathrm{added}} &=& \frac{\frac{c}{n}+v}{1+\frac{\frac{c}{n}v}{c^2}} \\   &=& \frac{c/n+v}{1+v/nc}  \\   &=& \frac{c(vn+c)}{nc+v} \end{eqnarray}$$

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Electric potential: Thin finite rod

Question:
A thin rod is placed along the z-axis that is stretched from $z=-d$ to $z=+d$. Let $\lambda$ be the linear density. There are two points $P_1 = (0,0,2d)$ and $P_2 = (x,0,0)$. Find the $x$ coordinate of $P_2$ so that potentials at $P_1$ and $P_2$ are equal.

Answer:
The potential at $P_1$ is positioned at the distance of $d$ from the edge of the rod, so the distance from the tiny piece of the charged rod should be $2d-z$. The charge is expressed by $\lambda dz$. Thus, the potential can be calculated as follows:
$$\begin{eqnarray} \phi_1 &=& \int \frac{kdq}{r} = k\int^{d}_{-d}\frac{\lambda dz}{2d-z}  \\   &=& -k\lambda \left[\ln(2d-z)\right]^{d}_{-d}  \\   &=& -k\lambda (\ln(2d-d)-\ln(2d+d))  \\   &=& -k\lambda (\ln(d)-\ln(3d))  \\   &=& -k\lambda \left(\ln\frac{1}{3}\right)  \\   &=& -k\lambda (-\ln(3)) \\   &=& k\lambda \ln(3) \end{eqnarray}$$
Likewise,
$$\begin{eqnarray} \phi_2 &=& k\int^{d}_{-d}\frac{\lambda dz}{\sqrt{x^2+z^2}}  \\   &=& k\lambda \ln\frac{\sqrt{x^2+d^2}+d}{\sqrt{x^2+d^2}-d} \end{eqnarray}$$
These potentials are equal, so
$$\begin{eqnarray} k\lambda \ln(3) &=& k\lambda \ln\frac{\sqrt{x^2+d^2}+d}{\sqrt{x^2+d^2}-d}  \\ 3 &=& \frac{\sqrt{x^2+d^2}+d}{\sqrt{x^2+d^2}-d}  \\ 3\sqrt{x^2+d^2}-3d &=& \sqrt{x^2+d^2}+d   \\ 2\sqrt{x^2+d^2} &=& 4d  \\ x^2 + d^2 &=& 4d^2  \\ x &=& \sqrt{3}d \end{eqnarray}$$

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Infinite integral with one constant and one variable

Question:
Find the value of
$$\int^{\infty}_{-\infty}e^{-x^2}\cos 2bxdx$$

Answer:
Let
$$\int^{\infty}_{-\infty}f(x,b)dx = 2\int^{\infty}_{0}f(x,b)dx$$
where $f(x,b)=e^{-x^2}\cos 2bx$. According to the properties of the functions, we can state that $|f(x,b)| \leqq e^{-x^2}$; and $\int^{\infty}_{0}e^{-x^2}$ is finite. Therefore, $\varphi(b)=\int^{\infty}_{0}f(x,b)dx$ is also finite. Now, take the partial derivative of the function $f(x,b)$ with respect to $b$.
$$f_b(x,b) = -2xe^{-x^2}\sin 2bx$$
This is continuous for arbitrary $b$ with $x \geqq 0$. Also we know $|f(x,b)| \leqq 2xe^{-x^2}$; and $\int^{\infty}_{0}2xe^{-x^2}$ is finite. Therefore, $\int^{\infty}_{0}f_b(x,b)dx$ is also finite for any $b$. Then,
$$\begin{eqnarray} \varphi'(b) &=& \int^{\infty}_{0}f_b(x,b)dx  \\   &=&  \int^{\infty}_{0}-2xe^{-x^2}\sin 2bxdx  \\    &=& \left[ e^{-x^2}\sin 2bx \right]^{\infty}_{0} - 2b\int^{\infty}_{0}e^{-x^2}\cos 2bxdx \\    &=& 0-2b\varphi(b) \end{eqnarray}$$
We have the following:
$$\begin{equation} \frac{\varphi'(b)}{\varphi(b)}=-2b \end{equation}$$
Then, integrate both sides.
$$\begin{eqnarray} \ln|\varphi(b)| &=& -b^2 + C  \\ \varphi(b) &=& Ce^{-b^2} \end{eqnarray}$$
From the initial condition,
$$\begin{equation} \varphi(0) = \int^{\infty}_{0}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}=C \end{equation}$$
Thus,
$$\begin{equation} \varphi(b)=\frac{\sqrt{\pi}}{2}e^{-b^2} \end{equation}$$
Therefore,
$$\begin{equation} \int^{\infty}_{-\infty}e^{-x^2}\cos 2bxdx = \sqrt{\pi}e^{-b^2} \end{equation}$$

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Electromagnetic waves: Maxwell's equations

Question:
The electric field of an electromagnetic wave is given as
$$\vec{E}=(E_{0x}\vec{x} + E_{0y}\vec{y})\sin(\omega t - kz + \phi)$$
Knowing that this is traveling along the z-axis. Find the magnetic field from this.

Answer:
From Maxwell's equations, we have
$$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$
This is known as Faraday's law. Calculate the left hand side.
$$\begin{eqnarray} \nabla \times \vec{E} &=& \left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)\vec{x}+\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)\vec{y} \\ &=& E_{0y}k\cos(\omega t-kz+\phi)\vec{x} - E_{0x}k\cos(\omega t-kz+\phi)\vec{y} \end{eqnarray}$$
This is equal to $-\frac{\partial \vec{B}}{\partial t}$. Thus, integrate both sides with respect to time to find the magnetic field.
$$\begin{eqnarray} \vec{B} &=& -\int (\nabla \times \vec{E}) dt  \\   &=& -E_{0y}\frac{k}{\omega}\sin(\omega t-kz+\phi)\vec{x} + E_{0x}\frac{k}{\omega}\sin(\omega t-kz+\phi)\vec{y}  \\   &=& (-E_{0y}\vec{x}+E_{0x}\vec{y})\frac{1}{c}\sin(\omega t - kz + \phi) \end{eqnarray}$$
Note that $\omega=2\pi f$ and $k=\frac{2\pi}{\lambda}$. Therefore, $\frac{k}{\omega}=\frac{1}{c}$ since $c=\lambda f$. The magnetic field propagates perpendicularly to the electric field with a shift of $\pi$ as you can see from the equations.

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Gravitational falling and kinematics

Question:
A stunt man jumps off from a 202-m high building onto a cushion having a thickness of 2.00 m. After his going into the cushion, it is crushed to a thickness of 0.500 m. What is the acceleration as he slows down?

Answer:
We need to find his velocity right before reaching the cushion. The cushion has 2.00-m high, so the stunt man falls by 200 m. The displacement, the initial velocity (0 m/s) and the acceleration (-9.80 m/s$^2$) are known. In order to find the final velocity, we use
$$\begin{eqnarray} v^2_f - v^2_0 &=& 2g\Delta h  \\ v_f &=& \sqrt{2g\Delta h}  \\       &=& \sqrt{2 \cdot (-9.8) \cdot (-200)}      &=& 62.61 \ \mathrm{m/s} \end{eqnarray}$$
Now, we calculate the deceleration of the man by the cushion using the same equation. The final velocity is zero and the initial velocity is 62.61 m/s. The final thickness is (0.500-2.00) m. Thus,
$$\begin{eqnarray} v^2_f - v^2_0 &=& 2ad  \\ 0 - 62.61^2 &=& 2a (0.5 - 2)  \\ a &=& \frac{-62.61^2}{2 \cdot (-1.5)}  \\ a &=& 1307 \ \mathrm{m/s}^2 \end{eqnarray}$$

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Galilean transformation

Question:
Physics laws cannot be altered by the observer's frames of reference. The transformation from a lab reference  frame to a moving reference frame is known as Galilean transformation. Show that Newton's laws are invariant under Galilean transformation.

Answer:
Galilean transformation subtracts the amount of displacement regarding velocity of the moving frame of reference to make the equivalent observation from the lab frame. Namely,
$$x' = x - vt, \quad t' = t$$
We assume that the time elapses equally for both frames; and the relative velocity is constant. Thus, the velocity becomes
$$v' = \frac{dx'}{dt'}=\frac{d}{dt}(x-vt)=\frac{dx}{dt}-v$$
The acceleration becomes
$$a' = \frac{d^2x'}{dt'^2}=\frac{dv'}{dt'}=\frac{d}{dt}\left(\frac{dx}{dt}-v \right)=\frac{d^2x}{dt^2}=a$$
Again, note that $v$ is a constant. Therefore, the Newton's equation of motion (laws) must be invariant under Galilean transformation.
$$F'=m\frac{d^2x'}{dt'^2}=ma=F$$
In other words, the physics law is the same from any observers.

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Gravity: Force and potential energy

Question:
Newtonian theory of gravity can be modified at short range. The potential energy between two objects is given as
$$U(r) = -\frac{Gmm'}{r}(1-ae^{-\frac{r}{\lambda}})$$
What is the force between $m$ and $m'$ for short distances ($r \ll \lambda$)?

Answer:
The force is conservative and calculated by the derivative of the potential energy with respect to the distance.
$$\begin{eqnarray} F &=& -\frac{dU}{dr}  \\  &=& -\frac{Gmm'}{r^2}(1-ae^{-\frac{r}{\lambda}})+\frac{Gmm'}{r}\frac{a}{\lambda}(1-ae^{-\frac{r}{\lambda}}) \\  &=& -\frac{Gmm'}{r^2}\left(1-ae^{-\frac{r}{\lambda}}\left(1+\frac{r}{\lambda}\right)\right) \end{eqnarray}$$
Since $r \ll \lambda$, $\frac{r}{\lambda}\approx 0$. Therefore, the force for short ranges is
$$F = -\frac{Gmm'}{r^2}(1-a)$$

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Magnetic fields from current flows and resistor circuit

Question:
The wire splits into two ways, which are bent circularly with radius of $a$. The upper half wire has resistance $2R$ and the lower part has $R$. Find the magnetic fields at the center of the circle in terms of the total current $I$.

Answer:
The magnetic field created by current $I_1$ is denoted as $B_1$. From Biot-Savart law,
$$\begin{eqnarray} \vec{B_1} &=& \oint \frac{\mu_0}{4\pi}\frac{I d\vec{l}\times \vec{r}}{r^2}  \\     &=& \frac{\mu_0}{4\pi}\frac{I_1}{a^2}\int_{0\rightarrow \pi a}dl (-\vec{z})  \\     &=& -\frac{\mu_0}{4\pi}\frac{I_1}{a^2}\pi a\vec{z} \\     &=& -\frac{\mu_0 I_1}{4a}\vec{z} \end{eqnarray}$$
Likewise, we can obtain the magnetic field created by the lower part.
$$\begin{equation} \vec{B_2} = \frac{\mu_0 I_2}{4a}\vec{z} \end{equation}$$
This is a parallel connection, and we can use
$$\frac{I_1}{I_2}=\frac{R}{2R} \quad \rightarrow \quad I_1=\frac{1}{2}I_2$$
The current is conserved.
$$I = I_1 + I_2 \quad \rightarrow \quad I_2 = I - I_1$$
Therefore,
$$I_1=\frac{1}{2}(I - I_1) \quad \rightarrow \quad I_1 = \frac{I}{3}$$
Hence, the other current will be
$$I_2 = \frac{2I}{3}$$
The magnetic field is expressed as
$$\vec{B}=\vec{B_1}+\vec{B_2}=\frac{\mu_0}{4a}(I_2-I_2)\vec{z}=\frac{\mu_0}{4a}\left(\frac{2I}{3}-\frac{I}{3}\right)\vec{z}=\frac{\mu_0 I}{12a}\vec{z}$$
The magnetic field is directed toward us.

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Matrix, eigenvalue, and eigenvector

Question:
(1) Find the eigenvalues and eigenvectors of a matrix,
$$A=\left( \begin{array}{c} -5 & 6  \\ -4 & 5 \end{array} \right)$$
(2) If $C$ is constructed by the eigen vectors obtained above, find whether $C$ is a regular function and whether $C^{-1}AC$ can be a diagonal matrix.

Answer:
(1) The eigenvalues can be obtained from $|\lambda E - A|=0$ where $E$ is the unit matrix.
$$\begin{eqnarray} |\lambda E - A| &=& \left| \begin{array}{c} \lambda+5 & -6  \\ 4 & \lambda-5 \end{array} \right| &=& (\lambda+5)(\lambda-5)+24  \\ &=& \lambda^2 -25 +24  \\ &=& \lambda^2 - 1 \end{eqnarray}$$
Therefore, the eigenvalues are $\lambda = \pm 1$. The eigen equation can be expressed as
$$A{\bf{x}} = \lambda{\bf{x}} \quad \mathrm{where} \ {\bf{x}}=\left( \begin{array}{c} x  \\ y \end{array} \right) \neq 0$$
Thus,
$$\begin{eqnarray*} \left( \begin{array}{c} -5-\lambda & 6 \\ -4 & 5-\lambda \end{array} \right) \left( \begin{array}{c} x  \\ y \end{array} \right) &=& 0   \\ \left( \begin{array}{c} -5-1 & 6 \\ -4 & 5-1 \end{array} \right) \left( \begin{array}{c} x  \\ y \end{array} \right) &=& \left( \begin{array}{c} -6 & 6 \\ -4 & 4 \end{array} \right) \left( \begin{array}{c} x  \\ y \end{array} \right) &=& 0 \end{eqnarray*}$$
Since the vector is not zero, in order to hold the equation, the ratio must be $x:y=1:1$. Likewise, plug in $\lambda=-1$; then, we have $x:y=3:2$.  Therefore, the eigen vectors are
$${\bf p}=\left( \begin{array}{c} x  \\ y \end{array} \right) = \left( \begin{array}{c} 1\\ 1 \end{array} \right), \quad {\bf q}=\left( \begin{array}{c} x  \\ y \end{array} \right) = \left( \begin{array}{c} 3 \\ 2 \end{array} \right)$$

(2) A matrix $C$ is combined by two eigen vectors obtained in question (1). Namely, $C=({\bf p},{\bf q})$.
$$C = \left(\begin{array}{c} 1 & 3 \\ 1 & 2 \end{array}\right)$$
Calculate the determinant.
$$\det C = 1 \cdot 2 - 1 \cdot 3 = -1$$
Since $\det \neq 0$, $C$ is regular. Now, calculate the following:
$$C^{-1}AC = \left(\begin{array}{c} -2 & 3 \\ 1 & -1 \end{array}\right) \left( \begin{array}{c} -5 & 6  \\ -4 & 5 \end{array} \right) \left(\begin{array}{c} 1 & 3 \\ 1 & 2 \end{array}\right) = \left( \begin{array}{c} 1 & 0  \\ 0 & -1 \end{array} \right)$$
This is a diagonal matrix and notice that the values correspond to the eigenvalues of matrix $A$.

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Reference,
The inverse matrix for a 2$\times$2 matix is
$$M = \left( \begin{array}{c} a & b  \\ c & d \end{array} \right), \quad M^{-1} = \frac{1}{\det M} \left( \begin{array}{c} d & -b  \\ -c & a \end{array} \right)$$

Horizontal spring motion: Wave equation

Question:
An object attached to a spring moves on a horizontal frictionless surface. This is a harmonic motion with amplitude of 0.16 m and period of 2.0 s. The mass is released from rest at $t=0$ s and $x=-0.16$ m. Find the displacement as a function of time.

Answer:
The simple harmonic motion is given as
$$x = A\cos(\omega t + \delta)$$
The amplitude, $A$, is 0.16 m. Since the period $T=2.0$ s, the frequency is calculated as
$$f = \frac{1}{2.0}=0.50 \ \mathrm{Hz}$$
Then, the angular frequency is
$$\omega = 2\pi f = \pi \ \mathrm{rad/s}$$
In order to find the phase, $\delta$, we need to check the initial state. When $t=0$, the wave equation becomes
$$\begin{eqnarray*} -0.16 &=& 0.16\cos(\delta)  \\ \cos(\delta) &=& -1.0  \\ \delta &=& \cos^{-1}(-1.0)  \\ \delta &=& \pi \end{eqnarray*}$$
Thus, we have the final form of the equation.
$$x = 0.16\cos(\pi t + \pi)$$

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RC circuit: With a DC power supply

Question:
Consider an RC circuit with a DC voltage supply as shown in the figure. There are four identical resistors ($R$=1.00 mega $\mathrm{\Omega}$) connected to one capacitor ($C$= 1.00 micro F). The voltage of the power supply is 10$\times$10$^6$ V. If the capacitor is fully charged and then the power supply is removed, find the current at $t=$0.5 s.

Answer:
All the four resistors are connected in parallel. The equivalent resistance of multiple parallel resistors is given by
$$\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_{\mathrm{1}}}+\frac{1}{R_{\mathrm{2}}}+\frac{1}{R_{\mathrm{3}}}+\cdots$$
In this case, we obtain
$$\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{4}{R}$$
Thus,
$$R_{\mathrm{eq}}=\frac{R}{4}=\frac{1.00\times 10^6}{4}=2.50\times 10^5 \mathrm{\Omega}$$
From Kirchoff's law, each circuit element consumes the voltage, and the total of it and supplied voltage must be zero. The capacitor consumes voltage, $\frac{Q}{C}$ where $Q$ is the total charge. The resistor consumes $RI$ because of Ohm's law. Since we consider the discharging process, voltage from the power supply is zero. Then, we have
$$-RI-\frac{Q}{C}=0$$
We know $I=\frac{dQ}{dt}$, so
$$\frac{dQ}{dt}+\frac{Q}{RC}=0$$
Solve this differential equation:
$$Q=Q_0 e^{-\frac{t}{RC}}$$
The charges and current flow are equivalent in terms of time, so can express it as
$$I=I_0 e^{-\frac{t}{RC}}$$
$RC$ is known as the time constant, $\tau=2.50\times 10^5 \times 1.00 \times 10^{-6}=0.25$ s. $I_0$ is the current right before the voltage source is removed; namely, the maximum current. It can be calculated by Ohm's law.
$$I_0=\frac{V}{R}=\frac{10\times 10^6}{2.50\times 10^5}=40.0 \ \mathrm{A}$$
Then, we can find the current at $t=0.5$ s.
$$I=40.0 e^{-\frac{0.5}{0.25}}=5.41 \ \mathrm{A}$$

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Vector calculus: Outer product and surface integral

Question:
A curved surface, $S$, is given as
$\vec{r}=\cos u \sin v \vec{i} + \sin u \sin v \vec{j} + \cos v \vec{k}$
$(0\leqq u \leqq 2\pi, 0\leqq v \leqq \pi)$
(1) Find $\frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v}$.
(2) Find $\iint_S \left| \frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v}\right| dudv$.

Answer:
(1) Take the partial derivative.
$$\begin{eqnarray} \frac{\partial\vec{r}}{\partial u} &=& -\sin u \sin v \vec{i} + \cos u \sin v \vec{j}  \\ \frac{\partial\vec{r}}{\partial v} &=& \cos u \cos v \vec{i} + \sin u \cos v \vec{j} -\sin v \vec{k} \end{eqnarray}$$
Therefore,
$$\frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v} = \left| \begin{array}{c} \vec{i}  &  \vec{j}  &  \vec{k}   \\ \left(\frac{\partial\vec{r}}{\partial u}\right)_i & \left(\frac{\partial\vec{r}}{\partial u}\right)_j & \left(\frac{\partial\vec{r}}{\partial u}\right)_k    \\ \left(\frac{\partial\vec{r}}{\partial v}\right)_i & \left(\frac{\partial\vec{r}}{\partial v}\right)_j & \left(\frac{\partial\vec{r}}{\partial v}\right)_k     \\ \end{array} \right|   \\ = \left| \begin{array}{c} \vec{i}  &  \vec{j}  &  \vec{k}   \\ -\sin u \sin v & \cos u \sin v & 0    \\ \cos u \cos v & \sin u \cos v & -\sin v     \\ \end{array} \right|   \\ = \left| \begin{array}{c} \cos u \sin v & 0    \\ \sin u \cos v & -\sin v     \\ \end{array} \right|\vec{i} - \left| \begin{array}{c} -\sin u \sin v & 0    \\ \cos u \cos v & -\sin v     \\ \end{array} \right| \vec{j} + \left| \begin{array}{c} -\sin u \sin v & \cos u \sin v     \\ \cos u \cos v & \sin u \cos v      \\ \end{array} \right|\vec{k}    \\ =-\cos u \sin^2 v\vec{i}-\sin u \sin^2 v\vec{j}-\sin v \cos v(\sin^2 u +\cos^2 u)\vec{k}  \\ =-\cos u \sin^2 v\vec{i}-\sin u \sin^2 v\vec{j}-\sin v \cos v \vec{k}$$

(2) Calculate $\left| \frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v}\right|$.
$$\begin{eqnarray*} \left| \frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v}\right| &=& \sqrt{|-\cos u \sin^2 v\vec{i}-\sin u \sin^2 v\vec{j}-\sin v \cos v \vec{k}|^2}  \\ &=& \sqrt{\cos^2 u \sin^4 v+\sin u^2 \sin^4 v+\sin^2 v \cos^2 v}   \\ &=& \sqrt{\sin^4 v(\cos^2 u+\sin u^2)+\sin^2 v \cos^2 v}   \\ &=& \sqrt{\sin^2 v(\sin^2 v+\cos^2 v)}   \\ &=& |\sin v| \end{eqnarray*}$$
Therefore, we have the integral.
$$\begin{eqnarray*} \int^{\pi}_0 |\sin v| dv \int^{2\pi}_0 du &=& [-\cos v]^{\pi}_0 \cdot 2\pi  \\      &=& [-\cos \pi -(-\cos 0)] \cdot 2\pi   \\      &=& [ 1 + 1] \cdot 2\pi  \\      &=& 4\pi \end{eqnarray*}$$

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Speed of waves: Water

Question:
The bulk modulus of water is 2.04$\times$10$^9$ Pa. The frequency of a water wave is found to be 262 Hz. What is the wavelength?

Answer:
The density of water is 1000 kg/m$^3$. The speed of the wave is given by
$$v = \sqrt{\frac{B}{\rho}}$$
where $B$ and $\rho$ are the bulk modulus and the density, respectively. Plug in the numbers.
$$v =  \sqrt{\frac{2.04\times 10^9}{1000}}=1428 \ \mathrm{m/s}$$
We know the relationship:
$$v = f\lambda$$
Therefore, the wavelength is
$$\lambda = \frac{v}{f}=\frac{1428}{262}=5.45 \ \mathrm{m}$$

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