Question

A particle of mass 2kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m . The height

Answer 1

Given:

Mass = 2 kg

radius = 0.6 m

\(w=12rad^{-1}\)

Distance (perpendicular) between point and mass, =

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

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

= \(m (wr)\eta \)

= 2 * 12 * 0.6 * 1

= 24 * 0.6 kg \(m^{2} s^{-1}\)

= 14.4 kg \(m^{2} s^{-1}\)