In Figure11.89depicts a device that can rotate freely with little friction with the axle. The radius is$\mathbf{0}\mathbf{.}\mathbf{4}\mathit{m}\mathbf{,}$and each of the eight balls has a mass of$\mathbf{0}\mathbf{.}\mathbf{3}\mathit{k}\mathit{g}\mathbf{.}$The device is initially not rotating. A piece of clay falls and sticks to one of the balls as shown in the figure. The mass of the clay is$\mathbf{0}\mathbf{.}\mathbf{066}\mathit{k}\mathit{g}$and its speed just before the collision is$\mathbf{10}\mathit{m}\mathbf{/}\mathit{s}\mathbf{.}$

(a) Which of the following statements are true, for angular momentum relative to the axle of the wheel? (1) Just before the collision, $\mathit{r}\mathbf{\perp }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\sqrt{\mathbf{2}}\mathbf{/}\mathbf{2}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{45}\mathbf{°}\mathbf{)}$(for the clay). (2) The angular momentum of the wheel is the same before and after the collision. (3) Just before the collision, the angular momentum of the wheel is$\mathbf{0}$. (4) The angular momentum of the wheel is the sum of the angular momentum of the wheel + clay after the collision is equal to the initial angular momentum of the clay. (6) The angular momentum of the falling clay is zero because the clay is moving in a straight line. (b) Just after the collision, what is the speed of one of the balls?

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.

The product of the moment of inertia and angular velocity is angular momentum, which is an important feature of a rotating object.

Use law of conservation of linear momentum and angular momentum to solve for speed of the one of the balls.

Figure shows the device containing the balls and their distances from the center of the device is given below.

Thus, the perpendicular distance of the clay just before the collision is $(0.4\sqrt{2}/2)m$and it is true.

(2) Because the wheel is at rest before to contact, the angular momentum of the wheel is initially zero. As a result, the stated conclusion is incorrect.

(3) Initially, the angular momentum of the wheel is zero, since the wheel is at rest before collision. Thus, the given statement is true.

(4) The angular momentum of a multi-particle system is equal to the sum of the angular momentum of all the particles in the system.

Thus, the given system is true.

(5) External force, and thus external torque, is zero in the wheel-clay system. As a result, the system's angular momentum will be conserved in the form of the given statement. As a result, the given assertion is correct.

(6) The formula for the angular momentum of the clay is,

$L_c=m_c{v}_c{r}_{\perp }$

For clay falling off the axle $v_c\ne 0$and localid="1668594635947" $r_{\perp }\ne 0$Thus,

$L_c\ne 0$

Hence, the given statement is False.

(b)Apply the law of conservation of angular momentum for the wheel clay system relative to axle.

The expression for the angular momentum of the clay-wheel system before collision is,

$L_{BWC}=m_C{v}_C{r}_{\perp }+I_w{\omega }_w$

Here, $I_w$is the moment of inertia of the wheel, and ${\omega }_w$is the initial angular speed of the wheel.

Substitute localid="1668594664043" $0.066kg$for localid="1668594670001" $10m/s$for localid="1668594738531" $v_C,(0.4\sqrt{2}/2)m$for localid="1668594650046" $r_{\perp }$, and 0 for localid="1668594680077" ${\omega }_w$in

$$\begin{gathered} L_{BWC}=m_C{v}_C{r}_{\perp }+I_w{\omega }_w \\ {L}_{BWC}=(0.066kg)(10m/s)(0.4\sqrt{2}/2m)+I_w\left(0\right) \\ =0.187kg.m^2/s \end{gathered}$$

Thus, the angular momentum of wheel, clay system before collision is $=0.187kg.m^2/s$

The angular momentum of the clay and wheel after collision is,

$L_{AWC}=(I_C+I_w)\omega$

Here, $I_C$is the moment of inertia of the clay, $I_w$is the moment of inertia of the wheel, and localid="1668594786213" $\omega$is the angular speed of rotation of the wheel.

The clay adheres to ball A, and the entire system revolves around the wheel's axel.

The phrase for the clay's moment of inertia is.

$I_C=m_c{r}^2$

Substitute localid="1668593840599">0.066kgfor $m_c,$and localid="1668594750779" $0.4m$for localid="1668594760696" $r$in $I_C=m_c{r}^2$

$$\begin{gathered} I_C=(0.066kg)(0.4m{)}^2 \\ =0.01056kg.m^2 \end{gathered}$$

The expression for the moment of inertia of the wheel is,

$I_w=8m{r}^2$

Substitute localid="1668594703490" $0.3kg$for localid="1668594773730" $m,$and $0.4m$for $r$in $I_W=8m{r}^2$

$$\begin{gathered} I_W=8(0.3kg)(0.4m{)}^2 \\ =0.384kg.m^2 \end{gathered}$$

Thus, the final angular momentum of the wheel-clay system is

$$\begin{gathered} L_{AWC}=(0.01056kg.m^2+0.384kg.m^2)\omega \\ =(0.395kg.m^2)\omega \end{gathered}$$

Conserve angular momentum of the clay-wheel system

$$\begin{gathered} L_{BWC}=L_{AWC} \\ =0.187kg.m^2/s=(0.395kg.m^2)\omega \\ =(0.395kg.m^2)\omega \end{gathered}$$

Thus, the angular velocity of the wheel is $0.473rad/s$.

The expression for the linear speed of the ball is,

$v=r\omega$

Substitute $0.473rad/s$for $\omega ,$and $0.4m$for $r$in$v=r\omega$

$$\begin{gathered} v=(0.4m)(0.473rad/s) \\ =0.189m/s \end{gathered}$$

Therefore, the speed of one of the balls of the wheel is $0.189m/s$.