Question

A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling what will be its speed when it starts pure rolling motion ?

Answer 1

Since there is no torque acting on the sphere with respect to contact point

So angular momentum will remain conserve

So initial angular momentum = final angular momentum

$muR = I_c\omega$

Momentum of inertia about the contact point will be given as

$I_c = \frac{7}{5}mR^2$

also at the time of pure rolling we have

$v = R\omega$

now again we have

$muR = \frac{7}{5}mR^2*\frac{v}{R}$

$muR = \frac{7}{5}mvR$

$v = \frac{5}{7}u$

so above is the speed of the center of sphere after it starts pure rolling

Answer 2

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