As shown in figure a uniform solid sphere rolls on a horizontal surface at 20 m/s. If then rolls up the incline shown. What will be the value of h, where the ball stops?

Question

As shown in figure a uniform solid sphere rolls on a horizontal surface at 20 m/s. If then rolls up the incline shown. What will be the value of h, where the ball stops? (A) 28.6m (B) 8.6m (C) 6m (D) 18.6m

Tardigrade Admin · Accepted Answer

The rotational and translational kinetic energy of the ball at the bottom will be changed to gravitational potential energy, when the sphere stops. We therefore write

(\frac{M v ^{2}}{2} + \frac{I ω ^{2}}{2})_{s t a r t} = (M g h)_{e n d}

For a solid sphere

I = \frac{2 M R ^{2}}{5}

Also,

ω = \frac{v}{r},

then above equation becomes

\frac{1}{2} M v^{2} + \frac{1}{2} (\frac{2}{5} M r^{2}) (\frac{v}{r})^{2} = M g h

\frac{1}{2} v^{2} + \frac{1}{5} v^{2} = (g + 8) h

Using v = 20 m/s gives h = 28.6 m

Q. As shown in figure a uniform solid sphere rolls on a horizontal surface at 20 m/s. If then rolls up the incline shown. What will be the value of h, where the ball stops?

28.6m

8.6m

6m

18.6m