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?

• A. $\frac{3u}{5}$
• B. $\frac{2u}{5}$
• C. $\frac{5u}{7}$
• D. $\frac{2u}{7}$

Answer 1

Answer: (C)

$\frac{5u}{7}$

Given,

$u=$ initial velocity of the solid sphere.

By the conservation of angular momentum,

$muR=I_c\omega$. . . . . . .(1)

Where, $R=$ radius of solid sphere

$\omega =$ angular velocity when sphere starts pure rolling motion.

The moment of inertia about the contact point will be given by

$I_c=\frac{7}{5}m{R}^2$ . . . . . (2)

When a solid sphere starts rolling motion then the speed is,

$v=\omega R$

$w=\frac{v}{R}$. . . . . . . . . . .(3)

Now, equation (1) becomes,

$muR=\frac{7}{5}m{R}^2\times \frac{v}{R}$

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

$v=\frac{5u}{7}m/s$

The correct option is C.