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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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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Potential energy and work: Spring

Question:
A spring is stretched when 4.0-kg object is hung on it vertically. The displacement is measured as 0.020 m. Then, one stretches it farther by 0.040 m. Find the work done by this external agent.

Answer:
Work can be obtained by integrating the force $|F|=kx$ with respect to correspondent displacement.
\begin{eqnarray}
W &=& \int^x_0 F\cdot dx  \\
  &=& \int^x_0 kx dx   = \frac{1}{2}kx^2
\end{eqnarray}
In order to calculate the work, we need to find the spring constant, $k$. Consider the free-body diagram. The spring and gravitational forces act on the hanging mass. According to the equation of motion, we have the net force:
\[
\sum F = kx - mg = ma = 0
\]
This system is in equilibrium, so $a=0$. Thus,
\[
kx = mg  \\
\rightarrow k = \frac{mg}{x} = \frac{4.0\cdot 9.8}{0.02}=1960 \ \mathrm{N/m}
\]
Now, we can calculate the work done by external agent.
\begin{eqnarray}
W &=& \frac{1}{2}kx^2  \\
    &=& \frac{1}{2}1960 \cdot 0.04^2  \\
    &=& 1.6 \ \mathrm{J}
\end{eqnarray}

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Newton's equation of motion

Question:
The objects A and B have masses, 20.0 kg and 10.0 kg, respectively. They are placed on a frictionless surface. The force (29.4 N) is acted as shown and these objects move together.
(1) Find the acceleration of the objects.
(2) What is the force exerting B from A?

Answer:
From the second figure, the forces exerting B from A and vice versa are equal due to the action and re-action pair.

(1) You can set up the Newton's equation for object A:
\[ \sum F_A = F - N = m_A a\]
The equation for object B is:
\[ \sum F_B = N = m_B a\]
Notice that the acceleration of each object is the same since they move together. Solve for the acceleration from them. Plug $N=m_B
a$ into $F - N = m_A a$.
\[ F - m_B a = m_A a \\
    (m_A + m_B)a = F  \\
   a = \frac{F}{m_A + m_B} \]
Therefore,
\[ a = \frac{29.4 \mathrm{N}}{20.0 \mathrm{kg} + 10.0 \mathrm{kg}} = 0.98 \mathrm{m/s^2}\]

(2) From the solution in (1), we can find the force exerting B from A, which is $N$ in the figure. Thus,
\[ N = m_B a = 10.0\mathrm{kg}\times 0.98\mathrm{m/s^2} = 9.8\mathrm{N}\]

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