Question

A thin uniform ring of mass M and radius R rotating
about its own axis with angular speed, is kept gently
on a rough horizontal surface. The coefficient of friction
between the ring and the surface is u. What will be
linear speed of ring when it starts pure rolling?
​

Answer 1

Answer:

The linear speed of ring when it starts pure rolling is

$v = \frac{R\omega_o}{2}$​

Explanation:

As we know that the friction force will act on the ring as kinetic friction from starting till it will perform pure rolling.

So here we can say that if reference is taken at the contact point of the ring and the ground then torque about that reference must be ZERO

So the total angular momentum must be conserved about that point.

So we will have

$L_i = L_f$

initial angular momentum is given as

$L_i = (MR^2)\omega_o$

final angular momentum is given as

$L_f = MvR + (MR^2)(\frac{v}{R})$

now from above equation

$(MR^2)\omega_o = 2MvR$

$v = \frac{R\omega_o}{2}$

#Learn

Topic : Angular Momentum conservation

https://brainly.in/question/9346398