Question

A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6m. The height of the table from the ground is 0.8m If the angular speed of the particle is $12rad{s}^{-1}$, the magnitude of its angular momentum about a point on the ground right under the centre of the circle is:

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

Answer 1

Answer: (C)

$14.4kg{m}^2{s}^{-1}$

Mass = 2 kg

radius = 0.6 m

$w=12ra{d}^{-1}$

Distance (perpendicular) between point and mass

$r_{\lambda }=\sqrt{{\left(0.6\right)}^2+{\left(0.8\right)}^2}=1m$

Angular momentum about the point on the ground under the centre of table = $mvr$

$\begin{array}{c}=m\left(wr\right)\eta \\ =2\times 12\times 0.6\times 1\\ =24\times 0.6kg{m}^2{s}^{-1}\\ =14.4kg{m}^2{s}^{-1}\end{array}$

Option C.