Question

A hoop of radius rand mass m rotating with an angular velocity w0 is placed on a rough horizontal surface. The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it ceases to slip?(a)w0r/4 (b)w0r/3 (c)w0r/2(d)wor

Answer 1

We know that,

$\bold{\omega = \frac{v}{r}}$

From conservation of momentum about bottom most point,

$\bold{mr^2\omega_0 = mvr+mr^2\times\frac{v}{r}}$

$\bold{\implies v= \frac{\omega_0r}{2}}$

Therefore,

Velocity of center of the hoop when it ceases to slip is   $\bold{\boxed{ v= \frac{\omega_0r}{2}}}$

Answer 2

hey...

(c) w0r/2

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