A rotating uniform-density disk of radius $\mathbf{0}\mathbf{.}\mathbf{6}\mathit{m}$is mounted in the vertical plane, as shown in Figure 11.88.The axle is held up by supports that are not shown, and the disk is free to rotate on the nearly frictionless axle. The disk has mass $\mathbf{5}\mathit{k}\mathit{g}\mathbf{.}$A lump of clay with mass$\mathbf{0}\mathbf{.}\mathbf{4}\mathit{k}\mathit{g}$falls and sticks to the outer edge of the wheel at the location $\mathbf{⟨}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{36}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{480}\mathbf{,}\mathbf{0}\mathbf{⟩}\mathit{m}\mathbf{,}$relative to an origin at the centre of the axle. Just before the impact the clay has speed $\mathbf{8}\mathit{m}\mathbf{/}\mathit{s}\mathbf{,}$and the disk is rotating clockwise with angular speed$\mathbf{0}\mathbf{.}\mathbf{51}\mathbf{}\mathbf{\text{radians/s.}}$

(a) Just before the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the center$\mathit{C}\mathbf{?}$(As usual,$\mathit{x}$is to the right,$\mathit{y}$is up, and$\mathit{z}$is out of the screen, toward you.) (b) Just after the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the center$\mathit{C}\mathbf{?}$(c) Just after the impact, what is the angular velocity (magnitude and direction) of the wheel? (d) Qualitatively, what happens to the linear momentum of the combined system? Why? (A) There is no change because linear momentum is always conserved. (B) Some of the linear momentum is changed into angular momentum. (C) Some of the linear momentum is changed into energy. (D) The downward linear momentum decreases because the axle exerts an upward force.

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

Radius of the uniform-density disk,$R=0.6m$

Mass of the disk,$M=5kg$

Mass of clay,$m=0.4kg$

Location of the clay,

Initial speed of the clay,$v=8m/s$

Initial angular speed of the disk,${\omega }_0=0.51rad/s$

A rotating uniform-density disk of radius $R$is mounted in the vertical plane is shown in the below figure.

In vector notation, initial velocity of the clay,

Initial angular velocity of the disk,

Initial linear momentum of the clay is,

Initial angular momentum of the clay is

Moment of inertia of the uniform-density disk of mass $M$and $R$is given by

$I=\frac{1}{2}M{R}^2$

Initial angular momentum of the disk can be calculated as

Therefore, the angular momentum of the combined system of wheel plus clay about the centre, just before impact is

The direction of the initial angular momentum of the system is out of the page.

The net external torque acting on the system is zero, so the angular momentum of the system is conserved.

$$\begin{gathered} {\overrightarrow{\tau }}_{net}=\frac{d\overrightarrow{L}}{dt} \\ 0=\frac{d\overrightarrow{L}}{dt} \\ {\overrightarrow{L}}_f={\overrightarrow{L}}_i \end{gathered}$$

Therefore, the angular momentum of the combined system of wheel plus clay about the centre, just after the impact is

The final moment of inertia of the system is equal to the sum of the moment of inertias of the clay and disk about the centre of the disk.

$$\begin{gathered} {\overrightarrow{I}}_f={\overrightarrow{I}}_{elay}+{\overrightarrow{I}}_{disk} \\ =m{R}^2+\frac{1}{2}M{R}^2 \\ =\left(m+\frac{M}{2}\right)R^2 \\ =\left(0.4kg+\frac{5kg}{2}\right)(0.6m{)}^2 \\ =1.044kg.m^2 \end{gathered}$$

The final rotational angular momentum of the system is

${\overrightarrow{L}}_f=I_f{\overrightarrow{\omega }}_f$

The final angular velocity of the wheel can be calculated as

Option D is correct, because the axle exerts an upward force in the direction opposite to the linear momentum.