Question

A circular disc of mass m and radius R is set into motion on horizontal floor with a linear speed v in the forward direction and an angular speed omega= v/R in clockwise direction. Find the magnitude of total angular momentum of the disc about bottommost point O of the disc......No spam answers plz​

Answer 1

Answer:

The magnitude of angular momentum about the lowest point will be

= I(about that axis through the bottom most point and perpendicular to the plane of disc) * Omega

Here, there won't any term accounting for the translational motion of the disc as the complete disc is performing pure rotation about the lowest point since the disc is pure rolling from the ground frame.

Hence,

L = (Icm + md^2)*v/R

L = (Icm + md^2)*v/R

L= (mR^2/2+mR^2)*v/R

Therefore,

L = 3mvR/2

So, to verify since we know that because the bottom most point of the disc while performing is actually at rest, which means that the angular momentum of the disc about this bottom most point and the angular momentum of the disc about any stationary point on the ground should be the same.

i.e 3mvR/2 which is true since,{ L = mR^2/2*v/R + mvR = 3mvR/2 }

Hope this helps you ! ｡◕‿◕｡