Question

Given ${V_{CM}} = 2\;{\rm{m/s}}$ $m = 2\;{\rm{kg}}$, $R = {\rm{4}}\;{\rm{m}}$.
Find angular momentum of the ring about the origin if it is in pure rolling.

A. \[32\;{\rm{kg}}{{\rm{m}}^2}/{\rm{s}}\]
B. $24\;{\rm{kg}}{{\rm{m}}^2}/{\rm{s}}$
C. $16\;{\rm{kg}}{{\rm{m}}^2}/{\rm{s}}$
D. $8\;{\rm{kg}}{{\rm{m}}^2}/{\rm{s}}$

Answer 1

Hint:We all know about the phenomenon of linear momentum, which is the multiplication of mass and velocity. Its harmonious form in angular motion is called angular momentum.So, it is the product of mass, speed, and radius of rotation. It is a conserved type of quantity.

Complete step by step answer:
We know that in the pure rolling condition, there is no sliding and slipping and hence the velocity of the center of mass for the body is represented as,
${V_{cm}} = R\omega $
Here ${V_{cm}}$ is the linear velocity of the center of mass of the body, and R is the radius of the body and $\omega $ is the angular velocity of the body.

We know that the angular momentum for a body about the origin is given as,
$L = m{V_{CM}}R + I\omega $
We will now substitute $\omega = \dfrac{{{V_{CM}}}}{R}$ , and $I = m{R^2}$ as this is the moment of inertia of the ring about the center, so we will get the result as,
\[
L = m{V_{CM}}R + m{R^2}\dfrac{{{V_{CM}}}}{R}\\
\Rightarrow L = m{V_{CM}}R + m{V_{CM}}R\\
\Rightarrow L = 2m{V_{CM}}R\\
\]
We will now substitute ${V_{CM}} = 2\;{\rm{m/s}}$ $m = 2\;{\rm{kg}}$, $R = {\rm{4}}\;{\rm{m}}$ to get the value of L.
\[
L = 2 \times \left( {2\;{\rm{kg}}} \right)\left( {2\;{\rm{m/s}}} \right)\left( {4\;{\rm{m}}} \right)\\
\therefore L{\rm{ = 32}}\;{\rm{kg - }}{{\rm{m}}^2}
\]
Here L is the angular momentum about the origin, v is the linear velocity of the body, and R is the radius of the body, and $\omega $ is the angular velocity of the body. Here ${V_{cm}}$ is the linear velocity of the center of mass of the body; R is the radius of the body and $\omega $ is the angular velocity of the body.

Therefore, the correct option is A.

Note:As we all have studied that linear momentum is a vector quantity, and its directions are the same towards the velocity vector; in the same way, angular momentum is also a vector quantity, and its direction is towards the direction of the angular velocity vector. If the linear momentum remains to conserve, then External forces acting on the body are zero. Still, if angular momentum is conserved, then external torque working on the body is zero.