Question

A uniform ring of radius r is given a backspin of angular velocity vo/2r and thrown on a horizontal rough surface with velocity of center to be vo. The velocity of centre of ring when it starts rolling motion will be

Answer 1

Final Answer : $\frac{v_o}{4} right$

Convention : Clockwise = positive, Anticlockwise = negative.

Steps:
1) We know, About point any point P on the ground ,Angular momentum will be conserved.
As torque due to friction blows off.

Let the final state has velocity 'v'and angular velocity $\omega$ .
Thus,
$v = \omega r$

2)

We have,
$I_{cm} = Mr^2 , v_i = v_o , v_f = v , \omega_i = \frac{-v_o}{2R} , \omega _f = \omega$

3) Then,
$\vec{L_P} = constant \\ \\ => mv_or -mr^2*\frac{v_o}{2r} = mvr + mr^2\omega \\ \\ => v_o - \frac{v_o}{2} = 2v \\ \\ => v = \frac{v_o}{4}$

Therefore, Final velocity of ring is
$v = \frac{v_o}{4}$
in forward direction.