Chapter 11, Problem 41P

(a)

To determine

The angular momentum of the bullet relative to the door’s axis of rotation.

Answer to Problem 41P

Yes, the bullet has angular momentum about the door fixing axis.

Explanation of Solution

Write the expression for angular momentum for an object as.

  $L=m_{B}r{v}_i$                                                                                                       (I)

Here, $L$ is the angular momentum of the object measured with respect to the axis of rotation for the object, $m_B$ is mass of the bullet, $v_i$ is the speed of bullet just before striking the door and $r$ is the distance between the bullets to door fixing axis.

Force acting on the door-bullet system as shown below.

Conclusion:

Yes, the bullet has angular momentum about the door. The value of angular momentum for bullet will be the product of the moment of inertia of bullet and the angular speed of the bullets.

Thus, yes the bullet has angular momentum about the door fixing axis.

(b)

To determine

The angular momentum for bullet relative to the door’s axis of rotation.

Answer to Problem 41P

The angular momentum for bullet relative to the door’s axis of rotation is $\underset{\_}{4.5\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}}$.

Explanation of Solution

Conclusion:

Substitute $0.0050\text{kg}$ for $m_B$, $1\times {10}^3\text{m}/\text{s}$ for $v_i$ and $0.900\text{m}$ for $r$ in equation (I).

  $\begin{array}{c}L=\left(0.0050\text{kg}\right)\left(0.900\text{m}\right)\left(1\times {10}^3\text{m}/\text{s}\right)\\ =4.5\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\end{array}$

Thus, the angular momentum for bullet relative to the door’s axis of rotation is $\underset{\_}{4.5\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}}$.

(c)

To determine

The mechanical energy of the bullet– door system during this collision.

Answer to Problem 41P

The mechanical energy of the bullet-door system during this collision will not remain constant.

Explanation of Solution

Conclusion:

The collision of bullet and door is an inelastic collision; therefore, the mechanical energy of the bullet door system will not remain constant during a collision. In an inelastic collision, there is high friction between bullet and door so the amount of mechanical energy or kinetic energy of the bullet converts into internal energy of the bullet door system.

Thus, the mechanical energy of the bullet-door system during this collision will not remain constant.

(d)

To determine

The angular speed at which the door swing open immediately after the collision.

Answer to Problem 41P

The angular speed at which the door swing open immediately after the collision is $\underset{\_}{0.749\text{rad}/\text{s}}$.

Explanation of Solution

Write the expression for moment of inertia of a bullet as.

  $I_b=m_B{r}^2$                                                                                                     (I)

Here, $I_b$ is the moment of inertia of a particle, $m_B$ is mass of the particle and $r$ is the radius of rotation.

Write the expression for moment of inertia for a stick or thin rod when it pivoted at another end as.

  $I_d=\frac{1}{3}M{D}^2$                                                                                             (II)

Here, $I_d$ the moment of inertia door, $M$ the mass of the door and $D$ is the length of the door.

Write the expression for total moment of inertia which is the sum of MOI’s of door and bullet as.

  $I=m_B{r}^2+\frac{1}{3}M{D}^2$                                                                                 (III)

Write the expression for total angular momentum as.

  $L=\left(m{r}^2+\frac{1}{3}M{D}^2\right)\omega$                                                                       (IV)

Here, $L$ is total angular momentum.

For conservation of angular momentum, equate the angular momentum of the bullet to the final angular momentum of door and bullet mass system as.

  $m_{B}r{v}_i=\left(m_B{r}^2+\frac{1}{3}M{D}^2\right)\omega$                                                                     (V)

Conclusion:

Substitute $0.0050\text{kg}$ for $m_B$, $1\times {10}^3\text{m}/\text{s}$ for $v_i$, $0.900\text{m}$ for $r$, $18.0\text{kg}$ for $M$ and $1.00\text{m}$ for $D$ in equation (V).

  $\left(0.0050\text{kg}\right)\left(0.900\text{m}\right)\left(1\times {10}^3\text{m}/\text{s}\right)=\left(\begin{array}{l}\left(0.0050\text{kg}\right){\left(0.900\text{m}\right)}^2\\ +\frac{1}{3}\left(18.0\text{kg}\right){\left(1.00\text{m}\right)}^2\end{array}\right)\omega$

Simplify the above expression for $\omega$ as.

  $\begin{array}{l}\omega =\frac{\left(0.0050\text{kg}\right)\left(0.900\text{m}\right)\left(1\times {10}^3\text{m}/\text{s}\right)}{\left(\left(0.0050\text{kg}\right){\left(0.900\text{m}\right)}^2+\frac{1}{3}\left(18.0\text{kg}\right){\left(1.00\text{m}\right)}^2\right)}\\ \omega =0.749\text{rad}/\text{s}\end{array}$

Thus, the angular speed at which the door swing open immediately after the collision is $\underset{\_}{0.749\text{rad}/\text{s}}$.

(e)

To determine

The kinetic energy of the door-bullet system immediately after impact.

Answer to Problem 41P

The kinetic energy of the door-bullet system immediately after impact $\underset{\_}{1.68\text{J}}$ and the kinetic energy of the bullet is $\underset{\_}{2.5\times {10}^3\text{J}}$.

Explanation of Solution

Write the expression for rotational kinetic energy as.

  ${\left(KE\right)}_r=\frac{1}{2}I{\omega }^2$                                                                                            (VI)

Here, ${\left(KE\right)}_r$ is rotational kinetic energy, $I$ is the moment of inertia and $\omega$ is angular speed.

Write the expression for kinetic energy for the bullet as.

  ${\left(KE\right)}_b=\frac{1}{2}{m}_B{v}_i{}^2$                                                                                        (VII)

Here, ${\left(KE\right)}_b$ is the kinetic energy of the bullet.

Substitute $m_B{r}^2+\frac{1}{3}M{D}^2$ for $I$ and $0.749\text{rad}/\text{s}$ for $\omega$ in equation (VI) then it gives the kinetic energy for the bullet-door system as.

  ${\left(KE\right)}_r=\frac{1}{2}\left(m_B{r}^2+\frac{1}{3}M{D}^2\right){\left(0.749\text{rad}/\text{s}\right)}^2$ (VIII)

Conclusion:

Substitute $0.0050\text{kg}$ for $m_B$, $0.900\text{m}$ for $r$, $18.0\text{kg}$ for $M$ and $1.00\text{m}$ for $D$ in equation (VIII).

  $\begin{array}{c}{\left(KE\right)}_r=\frac{1}{2}\left(\left(0.0050\text{kg}\right){\left(0.900\text{m}\right)}^2+\frac{1}{3}\left(18.0\text{kg}\right){\left(1.00\text{m}\right)}^2\right){\left(0.749\text{rad}/\text{s}\right)}^2\\ =1.68\text{J}\end{array}$

Substitute $0.0050\text{kg}$ for $m_B$, $1\times {10}^3\text{m}/\text{s}$ for $v_i$ in equation (VII).

  $\begin{array}{c}{\left(KE\right)}_b=\frac{1}{2}\left(0.0050\text{kg}\right){\left(1\times {10}^3\text{m}/\text{s}\right)}^2\\ =2.5\times {10}^3\text{J}\end{array}$

Thus, the kinetic energy of the door-bullet system immediately after impact $\underset{\_}{1.68\text{J}}$ and the kinetic energy of the bullet is $\underset{\_}{2.5\times {10}^3\text{J}}$.