Chapter 11, Problem 51AP

(a)

To determine

The appropriate analysis model to describe the projectile and the rod

Answer to Problem 51AP

The appropriate analysis model to describe the particle and the rod is a projectile motion and the rod behaves like an isolated system.

Explanation of Solution

Conclusion:

The mass is moving to the right with linear velocity, as it strikes to a stationary rod the mass will follow projectile motion with the rod. The other end of the rod will also rotate about the fixed point to balance the rod, so the angular momentum of the mass-rod system is conserved. The torque created by the mass and another end of the rod about the fix point $O$ is equal and opposite in direction, therefore, the net torque on the mass rod system becomes zero.

Thus, the appropriate analytical model to describe the particle and the rod is a projectile motion and the rod behaves like an isolated system.

(b)

To determine

The angular momentum of the system before the collision about an axis through $O$.

Answer to Problem 51AP

The angular momentum of the system before the collision about an axis through $O$ is $\underset{\_}{\frac{m{v}_{i}d}{2}}$.

Explanation of Solution

The angular momentum before the collision for a given system is the sum of angular momentum of ball and rod. But before the collision angular momentum of the rod is zero because the rod is stationary.

Consider the mass of the ball is concentrated at its centre so it behaves like a particle.

Write the expression for the angular momentum of the system before the collision as.

  $L_{t\text{otal}}=L_{\text{particle}}+L_{Rod}$

Substitute $0$ for $L_{Rod}$ in above expression as

  $L_{t\text{otal}}=L_{\text{particle}}+\left(0\right)$

Re-arrange the terms.

  $L_{t\text{otal}}=L_{\text{particle}}$                                                                                                 (I)

Here, $L_{t\text{otal}}$ total angular momentum for the given system before the collision with rod and $L_{\text{particle}}$ is angular momentum for the particle.

Write the expression for momentum for a particle as.

  $L_{particle}=\frac{m{v}_{i}d}{2}$                                                                                            (II)

Here, $m$ is mass of the particle, $v_i$ the initial speed of the particle and $d$ is the length of the rod.

Conclusion:

Substitute $\frac{m{v}_{i}d}{2}$ for $L_{particle}$ in equation (I)

  $L_{t\text{otal}}=\frac{m{v}_{i}d}{2}$

Thus, the angular momentum of the system before the collision about an axis through $O$ is $\underset{\_}{\frac{m{v}_{i}d}{2}}$.

(c)

To determine

The moment of inertia of the system about an axis through $O$ after the projectile sticks to the rod.

Answer to Problem 51AP

The moment of inertia of the system about an axis through $O$ after the projectile sticks to the rod is $\underset{\_}{\frac{\left(M+3m\right)d^2}{12}}$.

Explanation of Solution

The moment of inertia for a given system is the sum of the moment of inertia for particle and rod.

Write the expression for moment of inertia for a particle about the fix point $O$ as.

  $I_{particle}=m{\left(\frac{d}{2}\right)}^2$                                                                                          (III)

Here, $I_{particle}$ is moment of inertia of the particle about the fix point $O$.

Write the expression for moment of inertia for rod about the fix point $O$.

  $I_{Rod}=\frac{1}{12}M{d}^2$                                                                                             (IV)

Here, $I_{Rod}$ is moment of inertia of the rod about the fix point $O$.

Write the expression for total moment of inertia for rod-particle system about the fix point $O$ as.

  $I_{total}=I_{Rod}+I_{particle}$                                                                                      (V)

Here, $I_{total}$ is total moment of inertia of the rod-particle about the fix point $O$.

Substitute $\frac{1}{12}M{d}^2$ for $I_{Rod}$ and $m{\left(\frac{d}{2}\right)}^2$ for $I_{particle}$ in equation (V).

  $I_{total}=\frac{1}{12}M{d}^2+m{\left(\frac{d}{2}\right)}^2$                                                                             (VI)

Simplify the above equation for $I_{total}$ as.

  $\begin{array}{c}{I}_{total}=d^2\left(\frac{M}{12}+\frac{m}{4}\right)\\ =\frac{\left(M+3m\right)d^2}{12}\end{array}$

Conclusion:

Thus, the moment of inertia of the system about an axis through $O$ after the particle sticks to the rod is $\underset{\_}{\frac{\left(M+3m\right)d^2}{12}}$.

(d)

To determine

The angular momentum of the system after the collision.

Answer to Problem 51AP

The total angular momentum of the system after the collision is $\underset{\_}{\left(\frac{\left(M+3m\right)d^2}{12}\right)\omega }$.

Explanation of Solution

Write the expression for the angular momentum of the system after the collision.

  ${\left(L_{total}\right)}_f=I_{total}\omega$                                                                                         (VII)

Here, ${\left(L_{total}\right)}_f$ is the total angular momentum of the system after the collision and $\omega$ is angular velocity of the system after the collision.

Conclusion:

Substitute $\frac{\left(M+3m\right)d^2}{12}$ for $I_{total}$ in equation (VII).

  ${\left(L_{total}\right)}_f=\left(\frac{\left(M+3m\right)d^2}{12}\right)\omega$

Thus, the total angular momentum of the system after the collision is $\underset{\_}{\left(\frac{\left(M+3m\right)d^2}{12}\right)\omega }$.

(e)

To determine

The angular speed $\omega$ after the collision.

Answer to Problem 51AP

The angular speed $\omega$ after the collision is $\underset{\_}{\frac{6m{v}_i}{d\left(M+3m\right)}}$.

Explanation of Solution

Write the expression for conserved angular momentum as.

  ${\left(L_{total}\right)}_f=L_{total}$                                                                            (VIII)

Here, ${\left(L_{total}\right)}_f$ is final total angular momentum after the collision and $L_i$ is initial angular momentum for the particle.

Conclusion:

Substitute $\left(\frac{\left(M+3m\right)d^2}{12}\right)\omega$ for ${\left(L_{total}\right)}_f$ and $\frac{m{v}_{i}d}{2}$ for $L_{total}$ in equation (VIII).

  $\left(\frac{\left(M+3m\right)d^2}{12}\right)\omega =\frac{m{v}_{i}d}{2}$

Simplify the above expression for $\omega$ as.

  $\begin{array}{c}\omega =\frac{m{v}_{i}d}{2\left(\frac{\left(M+3m\right)d^2}{12}\right)}\\ =\frac{6m{v}_i}{d\left(M+3m\right)}\end{array}$

Thus, the angular speed $\omega$ after the collision is $\underset{\_}{\frac{6m{v}_i}{d\left(M+3m\right)}}$.

(f)

To determine

The kinetic energy of the system before the collision.

Answer to Problem 51AP

The kinetic energy of the particle before the collision is $\underset{\_}{\frac{1}{2}m{v}_i{}^2}$.

Explanation of Solution

The kinetic energy of the system before the collision is equal to kinetic energy of the particle because before the collision the rod is not in motion so the kinetic energy of the rod becomes zero. Therefore before the collision the kinetic energy of the system becomes the kinetic energy of the particle.

Write the expression for kinetic energy of the particle before the collision as.

  $K_{bc}=\frac{1}{2}m{v}_i{}^2$                                                                                                (IX)

Here, $K_{bc}$ is kinetic energy of the particle before the collision.

Conclusion:

Thus, the kinetic energy of the particle before the collision is $\underset{\_}{\frac{1}{2}m{v}_i{}^2}$.

(g)

To determine

The kinetic energy of the system after the collision

Answer to Problem 51AP

The kinetic energy of the system after the collision is $\underset{\_}{\frac{3m^2{v}_i{}^2}{2\left(M+3m\right)}}$.

Explanation of Solution

Write the expression for rotational kinetic energy of the particle-rod system after the collision as.

  $K_{total}=\frac{1}{2}{I}_{total}{\omega }^2$                                                                                            (X)

Here, $K_{total}$ is total rotational kinetic energy for the given system.

Conclusion:

Substitute $\frac{6m{v}_i}{d\left(M+3m\right)}$ for $\omega$ and $\left(\frac{\left(M+3m\right)d^2}{12}\right)$ for $I_{total}$ in equation (X).

  $\begin{array}{c}{K}_{total}=\frac{1}{2}\left(\frac{\left(M+3m\right)d^2}{12}\right){\left(\frac{6m{v}_i}{d\left(M+3m\right)}\right)}^2\\ =\left(\frac{\left(M+3m\right)d^2}{24}\right)\left(\frac{36m^2{v}_i{}^2}{d^2{\left(M+3m\right)}^2}\right)\end{array}$

Simplify the above expression as.

  $K_{total}==\frac{3m^2{v}_i{}^2}{2\left(M+3m\right)}$

Thus, the kinetic energy of the system after the collision is $\underset{\_}{\frac{3m^2{v}_i{}^2}{2\left(M+3m\right)}}$.

(h)

To determine

The fractional change of kinetic energy due to the collision.

Answer to Problem 51AP

The fractional change of kinetic energy due to the collision is $\underset{\_}{\frac{M}{M+3m}}$.

Explanation of Solution

Write the expression for change in kinetic energy for the system as.

  $\Delta K={\left(KE\right)}_{bc}-K_{total}$                                                                                   (XI)

Here, $\Delta K$ is change in kinetic energy of the system.

Write the expression for fractional change in kinetic energy as.

  $F_K=\frac{\Delta K}{K_{bc}}$                                                                                                   (XII)

Here, $F_K$ is fractional change in kinetic energy.

Conclusion:

Substitute $\frac{1}{2}m{v}_i{}^2$ for ${\left(KE\right)}_{bc}$ and $\frac{3m^2{v}_i{}^2}{2\left(M+3m\right)}$ for $K_{total}$ in equation (XI)

  $\begin{array}{c}\Delta K=\frac{1}{2}m{v}_i{}^2-\frac{3m^2{v}_i{}^2}{2\left(M+3m\right)}\\ =\frac{m{v}_i{}^2\left(M+3m\right)-3m^2{v}_i{}^2}{2\left(M+3m\right)}\\ =\frac{Mm{v}_i{}^2+3m^2{v}_i{}^2-3m^2{v}_i{}^2}{2\left(M+3m\right)}\end{array}$

Simplify the above expression for $\Delta K$ as.

  $\Delta K=\frac{mM{v}_i{}^2}{2\left(M+3m\right)}$

Substitute $\frac{mM{v}_i{}^2}{2\left(M+3m\right)}$ for $\Delta K$ and $\frac{1}{2}m{v}_i{}^2$ for $K_{bc}$ in equation (XII).

  $\begin{array}{c}{F}_K=\frac{\left(\frac{mM{v}_i{}^2}{2\left(M+3m\right)}\right)}{\frac{1}{2}m{v}_i{}^2}\\ =\frac{M}{M+3m}\end{array}$

Thus, the fractional change of kinetic energy due to the collision is $\underset{\_}{\frac{M}{M+3m}}$.