Question

A device consists of eight balls, each of massattached to the ends of low-mass spokes of length $\mathit{L}$ so the radius of rotation of ball is $\mathit{L}\mathbf{/}\mathbf{2}$. The device is mounted in the vertical plane, as shown in Figure 11.73. The axle is help up by supports that are not shown, and the wheel is free to rotate on the nearly frictionless axle. A lump of clay with mass$\mathit{m}$ falls and sticks to one of the balls at the location shown, when the spoke attached to that ball is $\mathbf{45}\mathbf{°}$to the horizontal. Just before the impact the clay has a speed $\mathit{v}$, and the wheel is rotating counter clock wise with angular speed$\mathit{\omega }$ .

(a.) Which of the following statements are true about the device and the clay, for angular momentum relative to the axle of the device? (1) the angular momentum of the device + clay just after the collision is equal to the angular momentum of the device +clay just before the collision. (2) The angular momentum of the falling clay is zero because the clay is moving in a straight line. (3) Just before the collision, the angular momentum of the wheel is 0. (4) The angular momentum of the device is the sum of the angular momenta of all eight balls. (5) The angular momentum of the device is the same before and after the collision. (b) Just before the impact, what is the (vector) angular momentum of the combined system of device plus clay about the center C? (As usual, $\mathit{x}$is to the right, $\mathit{y}$is up, and $\mathit{z}$is out of the screen, toward you) (c) Just after the impact, what is the angular momentum of the combined system of device plus clay about the center C? (d) Just after the impact, what is the (vector) angular velocity of the device? (e) Qualitatively. What happens to the total linear momentum is changed system? Why? (1) some of the linear momentum is changed into energy. (2) some of the linear momentum is changed into angular momentum. (3) There is no change because linear momentum is always conserved. (4) The downward linear momentum decreases because the axle exerts an upwards force. (f) qualitatively, what happens to the total kinetic energy of the combined system? Why? (1) some of the kinetic energy is changed into linear momentum. (2) some of the kinetic energy is changed into angular momentum. (3) The total kinetic energy decreases because there is an increase of internal energy in this inelastic collision. (4) There is no change because kinetic energy is always conserved.

Short answer

The total angular momentum of the device plus clay just before the collision is

$\left(2M\omega L^2-\frac{mvL}{2\sqrt{2}}\right)\hat{z}$

The net angular velocity of the device is$\left(\frac{2M\omega L^2-\frac{mvL}{2\sqrt{2}}}{2M\omega L^2-\frac{m{L}^2}{4}}\right)\stackrel{⏜}{z}$

Step-by-step solution

Definition of angular velocity.

The rate of change of angular displacement that describes the angular speed or rotational speed of an object and the axis around which the item is revolving is known as angular velocity. The amount of change in the particle's angular displacement over time is referred to as angular velocity.

The angular momentum of a rotational body is defined as the product of its moment of inertia and its angular velocity. In vector form, it is defined as the cross product of the particle position vector $\left(\overrightarrow{r}\right)$and its momentum vector \((\overrightarrow p )\)

Mathematically, it is represented as follows:

$L=I\omega =\overrightarrow{r}\times \overrightarrow{p}$

Here, $I$is the moment of inertia and $\omega$is the angular velocity.

Conservation of angular momentum states that the angular momentum of a system remains conserved unless it is acted upon by external torque. It is represented as follows:

$I\omega =$constant

Find the initial angular momentum of the device.

The total angular momentum of the device plus clay shortly after impact equals the angular momentum of the device plus clay just before the collision for the system, which is an axle and clay system.

Clay has a starting angular momentum of,

$L_{c1}=mv\left(\frac{L}{2}\right)\cos \theta$

Here,$m$ is the mass, $v$is the speed,$\frac{L}{2}$is the radius of rotation and$\theta$ is the angle between$\frac{L}{2}$and$mv$.

The initial angular momentum of device is,

$L_{d1}=I\omega$

Here,$I$is the total moment of inertia due to eight balls and${\omega }_i$is the initial angular speed.

Therefore, neither the angular momentum of falling clay is zero nor the angular momentum of wheel just before the collision is zero.

Thus, options (2) and (3) are incorrect.

The total moment of inertia due to eight balls

$I=8M{\left(\frac{L}{2}\right)}^2$

Here,$M$is the mass ball.

As a result, the device's angular momentum is equal to the sum of the angular momentum of all eight balls.

The device's angular momentum is not preserved because clay attaches to one of the balls, acting as an external torque. As there are no external torques acting on the system, the angular momentum of the device is not conserved, but the angular momentum of the device plus clay is conserved.

As a result, the proper choices are (1) and (2). (4).

Find the initial angular momentum of the clay.

The initial angular momentum of clay is,

$L_{c1}=mv\left(\frac{L}{2}\right)\cos \theta (-\hat{z}).....\left(1\right)$

Substitute $45°$ for $\theta$in equation (1)

$\begin{array}{rcl}{L}_{c1}& =& mv\left(\frac{L}{2}\right)\cos \theta (-\hat{z})\\ & =& \frac{mvL}{2}\cos \theta (-\hat{z})\\ & =& \frac{mvL}{2}\cos 45°(-\hat{z})\\ & =& \frac{mvL}{2\sqrt{2}}(-\hat{z})\\ & & \end{array}$

The initial angular momentum of device is,

$\begin{array}{rcl}{L}_{d1}& =& I{\omega }_i\left(\hat{z}\right)\\ & =& 8M{\left(\frac{L}{2}\right)}^2\left(\omega \right)\left(\hat{z}\right)\\ & =& 2M\omega L^2\left(\hat{z}\right)\\ & & \end{array}$

The total initial angular momentum of the system is,

$\begin{array}{rcl}{L}_i& =& L_{c1}+L_{d1}\\ & =& \frac{mvL}{2\sqrt{2}}(-\hat{z})+2M\omega L^2\left(\hat{z}\right)\\ & =& \left(2M\omega L^2-\frac{mvL}{2\sqrt{2}}\right)\hat{z}\\ & & \end{array}$

Thus, the total angular momentum of the device plus clay just before the collision is

$\left(2M\omega L^2-\frac{mvL}{2\sqrt{2}}\right)\hat{z}$

Find the total moment of inertia after the impact.

The angular momentum of the combined system is conserved before and after the collision.

$L_i=L_f=\left(2M\omega L^2-\frac{mvL}{2\sqrt{2}}\right)\hat{z}$

Therefore, the total angular momentum of the device plus clay just before the collision is

$\left(2M\omega L^2-\frac{mvL}{2\sqrt{2}}\right)\hat{z}$

The total moment of inertia after the impact is,

$\begin{array}{rcl}I& =& 8M{\left(\frac{L}{2}\right)}^2+m{\left(\frac{L}{2}\right)}^2\\ & =& 2M{L}^2+\frac{m{L}^2}{4}\\ & & \end{array}$

The net angular velocity of the device is,

$\begin{array}{rcl}{\omega }_r& =& \frac{L_f}{I}\\ & =& \left(\frac{2M\omega L^2-\frac{mvL}{2\sqrt{2}}}{2M\omega L^2-\frac{m{L}^2}{4}}\right)\stackrel{⏜}{z}\\ & & \end{array}$

Find the kinetic energy of system before collision.

The system's linear momentum remains constant. Linear and angular momentum conservation are independent of one another. Furthermore, momentum is never converted into energy.

As a result, both (2) and (1) are erroneous.

Initially, the clay ball has some angular momentum, which causes it to attach to the axle and conserve the system's angular momentum (axle plus clay).

As a result, there is no change in linear momentum.

As a result, option (3) is accurate, i.e. the total linear momentum of the combined system does not change because it is always conserved.

Before the impact, the system's kinetic energy,

$\begin{array}{rcl}K{E}_i& =& \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\\ & =& \frac{1}{2}m{v}^2+\frac{1}{2}\left(2M{L}^2\right){\omega }^2\\ & =& \frac{1}{2}m{v}^2+M{L}^2{\omega }^2\end{array}$

The kinetic energy of system after the collision,

$\begin{array}{rcl}K{E}_f& =& \frac{1}{2}I{\omega }_r^2\\ & =& \frac{1}{2}\left(2M{L}^2+\frac{m{L}^2}{4}\right){\left(\frac{2M\omega L^2-\frac{mvL}{2\sqrt{2}}}{2M{L}^2+\frac{m{L}^2}{4}}\right)}^2\\ & =& M{L}^2+\frac{m{L}^2}{8}{\left(\frac{2M\omega L^2-\frac{mvL}{2\sqrt{2}}}{2M{L}^2+\frac{m{L}^2}{4}}\right)}^2\\ & & \end{array}$

Therefore, the total kinetic energy of the system decreases.

Thus, the correct option is (3).