Question

A disc of mass m and radius R moves in the x-y plane as shown in the figure. The angular momentum of the disc at the instant shown is

• A. $\frac{5}{2}m{R}^2\omega$ about O
• B. $\frac{7}{2}m{R}^2\omega$ about O
• C. $\frac{1}{2}m{R}^2\omega$ about A
• D. $4m{R}^2\omega$ about A

Answer 1

Answer: (A), (C)

The correct options are
A

$\frac{5}{2}m{R}^2\omega$ about O

C

$\frac{1}{2}m{R}^2\omega$ about A

Angular momentum is the vector sum of two terms $I_{C_{\omega }}$ and $m{v}_{C{r}_1}$
Here, $I_{c_{\omega }}=\frac{1}{2}m{R}^2\omega$ (anticlockwise)
About O
$m{v}_{c{r}_1}=m\left(R\omega \right)\left(3R\right)$

$=3m{R}^2\omega$ (clockwise)
$\therefore L_{Total}=(3-\frac{1}{2})m{R}^2\omega$
$=\frac{5}{2}m{R}^2\omega$
About A
$r_1=0$
$\therefore m{v}_{C_{r_1}}=0$
or $L_{Total}=\frac{1}{2}m{R}^2\omega$