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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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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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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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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Ballistic pendulum

Question:
A 0.010-kg bullet is fired toward a 2.0-kg pendulum. The bullet is sucked in the pendulum block and both are raised up to 0.20-m height together. What is the initial velocity of the bullet?

Answer:
The masses of the bullet and the block can be denoted as $m$ and $M$, respectively. The initial velocity of the bullet and the final velocity of both the bullet and block are $v$ and $V$. Consider the states (a) through (b) in the figure. According to the conservation of linear momentum, the total initial momentum must be equal to the total final momentum.
\begin{equation}
mv = (m+M)V
\end{equation}
Now, consider the states (b) to (c). For those states, we use the conservation of mechanical energy.
\begin{equation}
\frac{1}{2}(m+M)V^2 = (m+M)gh
\end{equation}
$v$ and $V$ are unknown. In order to obtain the initial velocity, solve for $V$ first from the equation (1).
\begin{equation}
V = \frac{mv}{m+M}
\end{equation}
From equation (2), we have
\begin{equation}
V = \sqrt{2gh}
\end{equation}
Therefore,
\begin{equation}
\frac{mv}{m+M} = \sqrt{2gh}
\end{equation}
The initial velocity is
\begin{equation}
v = \frac{m+M}{m}\sqrt{2gh} = \frac{0.010+2.00}{0.010}\sqrt{2\cdot 9.8\cdot 0.20} = 398\quad\mathrm{m/s}
\end{equation}

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