Rolling Stones. A solid, uniform, spherical boulder starts from rest and rolls down a 50.0-m-high hill, as shown in Fig. P10.71. The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. What is the translational speed of the boulder when it reaches the bottom of the hill?

When one object’s surface comes in contact another one’s surface, the motion is resisted. This is known as Friction force.

When a body shifts from one point to another in space, it is known as translational motion. The speed corresponding to the translational motion is translational speed.

50 m high hill

So h = 25 m (top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction)

$$\begin{gathered} mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2 \\ \mathrm{Substituting}\mathrm{If}\mathrm{or}\mathrm{a}\mathrm{solid}\mathrm{sphere}\mathrm{and}\mathrm{using}\mathrm{v}=\mathrm{\omega r} \end{gathered}$$

$$\begin{gathered} mg\times \left(25m\right)=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2 \\ 50g=\frac{7}{10}{v}^2 \\ v=18.7\mathrm{m}/\mathrm{s} \end{gathered}$$
$$\begin{gathered} \mathrm{At}\mathrm{the}\mathrm{end}\mathrm{of}\mathrm{the}\mathrm{first}25\mathrm{m}\mathrm{the}\mathrm{boulder}\mathrm{has}\mathrm{a}\mathrm{velocity}\mathrm{of}18.7\mathrm{m}/\mathrm{s}. \\ \mathrm{In}\mathrm{the}\mathrm{second}25\mathrm{m}\mathrm{its}\mathrm{rotational}\mathrm{energy}\mathrm{does}\mathrm{not}\mathrm{change}\mathrm{but}\mathrm{its}\mathrm{translational}\mathrm{energy}\mathrm{does}. \end{gathered}$$

$$\begin{gathered} \mathrm{mgh}=△\mathrm{K} \\ \mathrm{mgh}=\frac{1}{2}\mathrm{m}\left({\mathrm{v}}_2^2-{\mathrm{v}}_1^2\right)\mathrm{where}\mathrm{h}=25\mathrm{m}\mathrm{and}{\mathrm{v}}_1=18.7\mathrm{m}/\mathrm{s} \\ \mathrm{g}\mathrm{x}25=\frac{1}{2}\left({\mathrm{v}}_2^2-18.7^2\right) \\ {\mathrm{v}}_2=28.98\mathrm{m}/\mathrm{s} \\ \approx 28\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the translational speed of the boulder is $\mathbf{29}\mathbf{m}\mathbf{/}\mathbf{s}$