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

Answer 1

Answer: (A)

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
$ {{\left( \frac{M{{v}^{2}}}{2}+\frac{I{{\omega }^{2}}}{2} \right)}_{start}}={{(Mgh)}_{end}} $
For a solid sphere
$ I=\frac{2M{{R}^{2}}}{5} $
Also, $ \omega =\frac{v}{r}, $ then above equation becomes
$ \frac{1}{2}M{{v}^{2}}+\frac{1}{2}\left( \frac{2}{5}M{{r}^{2}} \right){{\left( \frac{v}{r} \right)}^{2}}=Mgh $
$ \frac{1}{2}{{v}^{2}}+\frac{1}{5}{{v}^{2}}=(g+8)h $
Using v = 20 m/s gives h = 28.6 m