Question

A disc of mass M and radius R is rolling with an angular speed $\omega $on the horizontal plane. The magnitude of angular momentum of the disc about origin:
(A) $\dfrac{1}{2}M{R^2}\omega $
(B) $M{R^2}\omega $
(C) $\dfrac{3}{2}M{R^2}\omega $
(D) $2M{R^2}\omega $

Answer 1

Hint:The angular momentum is the rotational counterpart of linear momentum in translational motion which is defined as mass times velocity. Here the rotational analog of mass and velocity is the moment of inertia and angular velocity respectively. So,
\[
L = r \times p \\
L = I\omega \\
\]
Here the angular momentum is due to two motions viz, rotational motion of disc about its centre axis and angular momentum of linear motion of centre of mass about origin.

Complete step by step answer:
Just like we define linear momentum in translational motion, similarly in rotational motion we define angular momentum. Momentum origins from Newton’s 2nd Law of motion. It is defined as the mass times velocity or rotational motion moment of inertia times angular motion. It tells the quantity of motion in a body. It is logically opposite to inertia.
The quantity of motion of a body is needed to find the force which is required to keep in motion or the force to stop the motion and bring it to inertia. So, in rotational mechanics,
\[L = I\omega \]
Here two angular momentums are in play. Firstly, the angular momentum due to rotation about the centre of the disc and secondly the angular momentum due to the linear motion about the centre. So,
\[
\Rightarrow{L_{disc}} = I\omega \\
\Rightarrow I = \dfrac{1}{2}M{R^2} \\
\Rightarrow{L_{disc}} = \dfrac{1}{2}M{R^2}\omega \\
\]
The angular momentum about origin is therefore:
\[
\Rightarrow L = {L_{center}} + {L_{disc}} \\
\Rightarrow{L_{center}} = r \times p = M(R \times (R\omega )) \\
\Rightarrow{L_{center}} = M{R^2}\omega \\
\Rightarrow L = M{R^2}\omega + \dfrac{1}{2}M{R^2}\omega \\
\therefore L = \dfrac{3}{2}M{R^2}\omega \\
\]
The correct answer is Option C.

Note: Be careful when the angular momentum is asked. The point about which the angular momentum is asked is important. Here if the angular momentum was asked about the centre of disc then it would have been only L(disc). But the point origin was stated so we calculated the angular momentum due to the linear motion of its center of mass about the origin.