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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