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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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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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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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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Conservation of angular momentum

Question:
A solid cylinder with a moment of inertia, $I_0$, rotates about its center at an angular velocity of $\omega_0$. Another solid cylinder, which is initially rest, is put onto the rotating one gently. Both eventually rotate at an angular velocity, $\omega_f$. Find the velocity $\omega_f$.

Answer:
The angular momentum is defined as
\[
L = I\omega
\]
With the constant velocity, initial and final angular momenta are conserved. Namely, initial total momentum = final total momentum. In this case, we have
\[
I_0 \omega_0 = (I_0 + I_1)\omega_1
\]
Solve for $\omega_f$.
\[
\omega_f = \frac{I_0 \omega_0}{I_0 + I_1}
\]

Torques and the balance

Question:
A horizontal beam is attached to a wall. The length and weight are 10.0 m and 200 N, respectively. The far end is supported by a wire and the angle between the beam and wire is 60$^o$. A person who has 500-N weight stands 2.00 m from the wall. What is the tension in the wire?

Answer:
This system is at equilibrium, so the sum of all of the torques must be zero.
\[
\sum \tau = 0
\]
The torque is defined as
\[
\tau = rF\sin \theta
\]
where $r$ and $F$ are the lever arm and the force. The angle $\theta$ is taken from the beam. The weight of the beam acts on the middle of the length by assuming that the mass is uniformly distributed. Plug in the numbers to calculate the net torque:
\begin{eqnarray*}
500\cdot 2.00 \sin(270^o) &+& 200\cdot (5.00)\sin(270^o) + T\cdot 10.0 \sin(60^o) = 0    \\
- 1000 - 1000 &+& 10.0T\sin(60^o)  = 0     \\
T &=& \frac{2000}{10\sin(60^o)}       \\
T &=& 231 \quad \mathrm{N}
\end{eqnarray*}

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Lagrangian function of a sphere on a cylindrical surface

Question:
There is an open cylinder of radius $R$., which is stationary. A hollow sphere of radius $\rho$ and mass $m$ rolls on the surface without slipping. Find the Lagrangian function.

Answer:
The Lagrangian function is defined as
\[ L = KE - PE \]
where $KE$ and $PE$ represent kinetic energy and potential energy, respectively.
The mechanical potential energy is the gravitational force $\times$ height, which is taken from the center. The height varies with the angle $\theta$. Thus,
\[ PE = -mg(R-\rho)\cos\theta \]
The negative sign indicates the height is directed to the center.
Now the kinetic energy has two parts, translational and rotational kinetic energies.
\[ KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
The translational velocity, $v$, is along with the trace of circle whose radius is $R-\rho$. Therefore,
\[ v = (R-\rho)\omega \]
The angular velocity can be expressed by
\[ \omega = \theta' = \frac{d\theta}{dt} \]
The moment of inertia of hollow sphere is
\[ I = \frac{2}{3}m\rho^2 \]
Plug everything into above.
\[ L = \frac{1}{2}m(R-\rho)^2 \theta '^2 + \frac{1}{2}\frac{2}{3}m\rho^2 \theta '^2 + mg(R-\rho)\cos\theta \]
We can rewrite it as
\[ L = \frac{m\theta'^2}{6}(3R^2 -6R\rho + 5\rho^2)+ mg(R-\rho)\cos\theta \]

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Mechanical model of the proton (angular momentum etc.)

Question:
The proton's mass and radius are 1.67 $\times$ 10$^{-27}$ kg and 1.00 $\times$ 10$^{-15}$ m. This system is governed by quantum and classical regimes. Namely, the proton is assumed to have a rotational uniform solid spherical body and a half spin of quantum angular momentum. Find the equatorial velocity of the proton.

Answer:
In terms of classical sense, the moment of inertia of proton is
\[ I = \frac{2}{5}mr^2 \]
Therefore, the angular momentum becomes
\[ L = I\omega = \frac{2}{5}mr^2\omega \]
For the quantum perspective, the spin angular momentum is given as
\[ S = \frac{1}{2}\hbar \]
Therefore,
\[ \frac{2}{5}mr^2\omega  = \frac{1}{2}\hbar \]
Solve for the angular velocity.
\[ \omega = \frac{1}{2}\hbar \frac{5}{2mr^2} \]
The equatorial velocity is given by
\[ v = r\omega =  \frac{5\hbar}{4mr} \\
      = \frac{5}{4}\frac{1.055\times 10^{-34}\mathrm{Js}}{1.673\times 10^{-27}\mathrm{kg}\times 1.00\times 10^{-15}\mathrm{m}} \]
Then we have
\[ v = 7.88 \times 10^7 \mathrm{m/s} \]

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