Chapter 10, Problem 81P

A projectile of mass m moves to the right with a speed v_i (Fig. P10.81a). The projectile strikes and sticks to the end of a stationary rod of mass M, length d, pivoted about a frictionless axle perpendicular to the page through O (Fig. P10.81b). We wish to find the fractional change of kinetic energy in the system due to the collision. (a) What is the appropriate analysis model to describe the projectile and the rod? (b) What is the angular momentum of the system before the collision about an axis through O? (c) What is the moment of inertia of the system about an axis through O after the projectile sticks to the rod? (d) If the angular speed of the system after the collision is ω, what is the angular momentum of the system after the collision? (e) Find the angular speed ω after the collision in terms of the given quantities. (f) What is the kinetic energy of the system before the collision? (g) What is the kinetic energy of the system after the collision? (h) Determine the fractional change of kinetic energy due to the collision.

Figure P10.81

To determine

The appropriate model to analyze the system.

Answer to Problem 81P

The appropriate model to analyze the system is by considering it as an $\underset{\_}{\text{Isolated system}}$.

Explanation of Solution

The striking and sticking of the given projectile on the stationary rod can be considered as a collision. The collision occurring between two object is an isolated system for which the total momentum is conserved. The momentum of both objects before and after collision will be same. This is because the system is free from any external force which changes the momentum.

Since the rod and projectile is not experiencing any external force and torque the total momentum of the system will be conserved, and the system can be considered as isolated. Thus, the best suited analysis model is by treating the system as isolated.

Conclusion

Therefore, the appropriate model to analyze the system is by considering it as an $\underset{\_}{\text{Isolated system}}$.

To determine

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

Answer to Problem 81P

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

Explanation of Solution

The total angular momentum is the sum of the angular momentum of projectile and the rod. Since the rod is initially at rest its angular momentum before collision will be zero.

Write the expression for the total angular momentum.

  $L_{\text{totlal}}=L_{\text{particle}}+L_{\text{rod}}$        (I)

Here, $L_{\text{particle}}$ is the angular momentum of the projectile, and $L_{\text{rod }}$ is the angular momentum of the rod.

Write the expression for the angular momentum of the projectile at $O$.

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

Here, $m$ is the mass of projectile, $d$ is the length of the rod, ${\upsilon }_i$ is the initial speed of the projectile.

Conclusion:

Substitute, $\frac{m{\upsilon }_{i}d}{2}$ for $L_{\text{particle}}$, and $0$ for $L_{\text{rod }}$ in equation (I) to find the angular momentum of the system before collision.

  $\begin{array}{c}{L}_{\text{totlal}}=\frac{m{\upsilon }_{i}d}{2}+0\\ =\frac{m{\upsilon }_{i}d}{2}\end{array}$

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

To determine

The moment of inertia of the rod after collision.

Answer to Problem 81P

The moment of inertia of the rod after collision is $\underset{\_}{\left(\frac{d^2\left(M+3m\right)}{12}\right)}$.

Explanation of Solution

The total moment of inertia is the sum of the moment of inertia of rod and the projectile.

Write the expression for the total moment of inertia.

  $I_{\text{total}}=I_{\text{particle}}+I_{\text{rod}}$        (III)

Here, $I_{\text{particle }}$ is the moment of inertia of the projectile, and $I_{\text{rod}}$ is the moment of inertia of the rod.

Let $O$ can be considered as the center of mass of the rod. The moment of inertia of the rod about the center of mass is.

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

Here, $M$ is the mass of the rod.

Write the expression for the moment of inertia of the projectile about an axis passing through $O$.

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

Conclusion:

Substitute, equation (IV) and (V) in (III).

  $\begin{array}{c}{I}_{\text{total}}=\frac{1}{12}M{d}^2+m{\left(\frac{d}{2}\right)}^2\\ =\frac{d^2\left(M+3m\right)}{12}\end{array}$

Therefore, the moment of inertia of the rod after collision is $\underset{\_}{\left(\frac{d^2\left(M+3m\right)}{12}\right)}$.

To determine

The angular momentum of the system after collision.

Answer to Problem 81P

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

Explanation of Solution

After the collision there is only a single angular momentum since the projectile stick to the rod after striking.

Write the expression for the final angular momentum.

  $L_{\text{total}}=I_{\text{total}}\omega$        (VI)

Here, $\omega$ is the angular speed of the system.

Conclusion:

Substitute, $\frac{d^2\left(M+3m\right)}{12}$ for $I_{\text{total}}$ in equation (VI) to find the angular momentum of the system after collision.

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

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

To determine

The angular speed after the collision.

Answer to Problem 81P

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

Explanation of Solution

According to principle of conservation of angular momentum the momentum after and before collision will be same.

  $L_f=L_i$        (VII)

Conclusion:

Substitute, $\left(\frac{d^2\left(M+3m\right)}{12}\right)\omega$ for $L_f$, and $\frac{m{\upsilon }_{i}d}{2}$ for $L_i$ in equation (VII) and rearrange to obtain an expression for $\omega$.

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

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

To determine

The kinetic energy of the system before collision.

Answer to Problem 81P

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

Explanation of Solution

Since the rod is at rest the kinetic energy is only for the projectile. The projectile has mass $m$, and the speed of the projectile is ${\upsilon }_i$.

Hence the kinetic energy of the projectile is.

  $K=\frac{1}{2}m{\upsilon }_i^2$

Conclusion:

Therefore, the kinetic energy of the system before collision is $\underset{\_}{\frac{1}{2}m{\upsilon }_i^2}$.

To determine

The kinetic energy of the system after collision

Answer to Problem 81P

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

Explanation of Solution

The kinetic energy after the collision is the rotational kinetic energy of the system.

Write the expression for the rotational kinetic energy.

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

Conclusion:

Substitute, $\frac{6m{\upsilon }_i}{d\left(M+3m\right)}$ for $\omega$, and $\left(\frac{d^2\left(M+3m\right)}{12}\right)$ for $I_{\text{total}}$ in equation (VIII) to find the kinetic energy after the collision.

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

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

To determine

The fractional change in kinetic energy due o collision.

Answer to Problem 81P

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

Explanation of Solution

The change in energy is obtained by taking the difference of energy before, and after collision.

Write the expression for change in kinetic energy.

  $\left|\Delta K\right|=K_i-K_f$        (IX)

Substitute, $\frac{1}{2}m{\upsilon }_i^2$ for $K_i$, and $\frac{3m^2{\upsilon }_i^2}{2\left(M+3m\right)}$ for $K_f$ in equation (IX).

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

Write the expression for the fractional change in kinetic energy.

  $K_{\text{fractional}}=\frac{\left|\Delta K\right|}{K_i}$        (X)

Conclusion:

Substitute, $\frac{1}{2}m{\upsilon }_i^2$ for $K_i$, and $\frac{mM{\upsilon }_i^2}{2\left(M+3m\right)}$ for $\left|\Delta K\right|$ in equation (X) to find the fractional change in kinetic energy.

  $\begin{array}{c}{K}_{\text{fractional}}=\frac{\frac{mM{\upsilon }_i^2}{2\left(M+3m\right)}}{\frac{1}{2}m{\upsilon }_i^2}\\ =\frac{M}{M+3m}\end{array}$

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