Question

A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad/s, the magnitude of its angular momentum about a point on the ground right under the centre of the circle is

• A. $14.4kg{m}^2{s}^{-1}$
• B. $11.52kg{m}^2{s}^{-1}$
• C. $20.16kg{m}^2{s}^{-1}$
• D. $8.64kg{m}^2{s}^{-1}$

Answer 1

The correct option is A $14.4kg{m}^2{s}^{-1}$

$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}=r\cos \theta \hat{i}+r\sin \theta \hat{j}+z\hat{k}=0.6(\cos \theta \hat{i}+\sin \theta \hat{j})+0.8\hat{k}$ where $\theta$ is the angle the radius makes with the x-axis as the particle moves in the circle.
$\overrightarrow{p}=-m\left(r\omega \right)(\sin \theta \hat{i}+\cos \theta \hat{j})$
$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}=\left(0.6\right(\cos \theta \hat{i}+\sin \theta \hat{j})+0.8\hat{k})\times (-mv(\sin \theta \hat{i}+\cos \theta \hat{j}\left)\right)=2\times \left(0.6\omega \right)$
$=2\times 0.6\times 12$
$=14.4kg{m}^2/s$