A sphere and a cylinder have the same radius and the same mass. They start from rest at the top of an incline. (a) Which reaches the bottom first? (b) Which has the greater speed at the bottom? (c) Which has the greater total kinetic energy at the bottom? (d) Which has the greater rotational kinetic energy? Explain your answers.

The object’s angular velocity can be calculated by dividing the object’s linear velocity by the radius of the rounded path. Its value is altered inversely to the radius of the rounded path.

(a)

Let the mass of the sphere and cylinder be\(M\)and radius be\(R\).

The expression for the moment of inertia of the sphere is as follows:

\({I_{\rm{S}}} = \frac{2}{5}M{R^2}\)

The expression for the moment of inertia of the cylinder is as follows:

\({I_{\rm{C}}} = \frac{1}{2}M{R^2}\)

The expression for the angular velocity of the sphere is as follows:

\({\omega _{\rm{S}}} = \frac{{{v_{\rm{S}}}}}{R}\)

The expression for the angular velocity of the cylinder is as follows:

\({\omega _{\rm{C}}} = \frac{{{v_{\rm{C}}}}}{R}\)

For the sphere, apply the conservation of energy principle.

\(\begin{aligned}MgH &= \frac{1}{2}Mv_{\rm{S}}^2 + \frac{1}{2}{I_{\rm{S}}}\omega _{\rm{S}}^2\\MgH &= \frac{1}{2}Mv_{\rm{S}}^2 + \frac{1}{2}\left( {\frac{2}{5}M{R^2}} \right){\left( {\frac{{{v_{\rm{S}}}}}{R}} \right)^2}\\MgH &= \frac{1}{2}Mv_{\rm{S}}^2 + \frac{2}{{10}}Mv_{\rm{S}}^2\\{v_{\rm{S}}} &= \sqrt {\frac{{10gH}}{7}} \end{aligned}\) … (i)

For the cylinder, apply the conservation of energy principle.

\(\begin{aligned}MgH &= \frac{1}{2}Mv_{\rm{C}}^2 + \frac{1}{2}{I_{\rm{C}}}\omega _{\rm{C}}^2\\MgH &= \frac{1}{2}Mv_{\rm{C}}^2 + \frac{1}{2}\left( {\frac{1}{2}M{R^2}} \right){\left( {\frac{{{v_{\rm{C}}}}}{R}} \right)^2}\\MgH &= \frac{1}{2}Mv_{\rm{C}}^2 + \frac{1}{4}Mv_{\rm{C}}^2\\{v_{\rm{C}}} &= \sqrt {\frac{{4gH}}{3}} \end{aligned}\) … (ii)

From equations (i) and (ii), it is clear that\({v_{\rm{S}}} > {v_{\rm{C}}}\).

Thus, the sphere will reach the bottom first.

(b)

Thus, the sphere will have a greater speed at the bottom.

(c)

The initial potential energy of the sphere and the cylinder is similar at the top of the incline. Therefore, both have the same energy at the bottom.

(d)

The expression for the rotational kinetic energy of the sphere is as follows:

\(\begin{aligned}K{E_{\rm{S}}} &= \frac{1}{2}{I_{\rm{S}}}\omega _{\rm{S}}^2\\K{E_{\rm{S}}} &= \frac{1}{2}\left( {\frac{2}{5}M{R^2}} \right){\left( {\frac{{{v_{\rm{S}}}}}{R}} \right)^2}\\K{E_{\rm{S}}} &= \frac{1}{5}Mv_{\rm{S}}^2\\K{E_{\rm{S}}} &= \frac{1}{5}M\left( {\frac{{10gH}}{7}} \right)\\K{E_{\rm{S}}} &= \frac{{2MgH}}{7}\end{aligned}\) … (iii)

The expression for the rotational kinetic energy of the cylinder is as follows:

\(\begin{aligned}K{E_{\rm{C}}} &= \frac{1}{2}{I_{\rm{C}}}\omega _{\rm{C}}^2\\K{E_{\rm{C}}} &= \frac{1}{2}\left( {\frac{1}{2}M{R^2}} \right){\left( {\frac{{{v_{\rm{C}}}}}{R}} \right)^2}\\K{E_{\rm{C}}} &= \frac{1}{4}Mv_{\rm{C}}^2\\K{E_{\rm{C}}} &= \frac{1}{4}M\left( {\frac{{4gH}}{3}} \right)\\K{E_{\rm{C}}} &= \frac{{MgH}}{3}\end{aligned}\) … (iv)

Thus, from equations (iii) and (iv), it is clear that\(K{E_{\rm{C}}} < K{E_{\rm{S}}}\).

Hence, the cylinder will have the greater rotational kinetic energy at the bottom.