Chapter 8, Problem 14Q

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.

To determine

Between the sphere and the cylinder

1. The one that reaches the bottom first.
2. The one that has the greater speed at the bottom.
3. The one that has the greater total kinetic energy at the bottom.
4. The one that has the greater rotational kinetic energy.

Answer to Problem 14Q

Solution:

(a) The sphere reaches the bottom first.

(b) The sphere has the greater speed at the bottom.

(c) The total kinetic energy is the same for both at the bottom.

(d) The rotational kinetic energy of the cylinder is greater than that of the sphere.

Explanation of Solution

Given:

The sphere and the cylinder have the same radius ( r ) and the same mass ( m )

Formula used:

For a sphere, moment of inertia is $I_s=\left(\frac{2}{5}\right)m{r}^2$

For a cylinder, moment of inertia is $I_c=\frac{1}{2}m{r}^2$

Potential energy:

  $PE=mgh$

Total kinetic energy:

  $KE=K{E}_{translational}+K{E}_{rotational}$

Angular speed:

  $\omega =\frac{v}{r}$

Here, m is the mass, r is the radius, g is the gravitational acceleration, h is the height and v is the linear velocity.

Calculation:

The potential energy of the sphere as well as the cylinder at the top of the incline is given by $PE=mgh$ , where h is the height of the incline

At the bottom of the incline, this PE gets converted entirely into kinetic energy ( KE ), which is given by

  $KE=K{E}_{translational}+K{E}_{rotational}$

  $\Rightarrow KE=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Where, v is the speed of the object at the bottom of the incline, ω is the angular velocity of the rolling object and I is the moment of inertia

For the sphere, $I_s=\left(\frac{2}{5}\right)m{r}^2$

For the cylinder, $I_c=\frac{1}{2}m{r}^2$

Also, $\omega =\frac{v}{r}$

Therefore, $K{E}_{sphere}=\frac{1}{2}m{v}_s{}^2+\frac{1}{2}\times \frac{2}{5}m{r}^2\times {\left(\frac{v_s}{r}\right)}^2$

  $K{E}_{sphere}=\frac{1}{2}m{v}_s{}^2+\frac{1}{5}m{v}_s{}^2=\frac{7}{10}m{v}_s{}^2$

  $K{E}_{cylinder}=\frac{1}{2}m{v}_s{}^2+\frac{1}{2}\times \frac{1}{2}m{r}^2\times {\left(\frac{v_s}{r}\right)}^2$

  $\begin{array}{l}K{E}_{cylinder}=\frac{1}{2}m{v}_s{}^2+\frac{1}{4}m{v}_s{}^2\\ K{E}_{cylinder}=\frac{6}{8}m{v}_s{}^2\end{array}$

For the sphere, $mgh=\frac{7}{10}m{v}_s{}^2$

or $v_s=\sqrt{\frac{10}{7}gh}$ ---(i)

For the cylinder, $mgh=\frac{6}{8}m{v}_c{}^2$

or $v_c=\sqrt{\frac{4}{3}gh}$ ---(ii)

From (i) and (ii), $v_s>v_c$

Therefore, the sphere reaches the bottom first and has the greater speed at the bottom

Since, the initial potential energy is equal for the two objects, the total kinetic energy at the bottom of the incline is also the same for both.

  $K{E}_{rotational/sphere}{}_{}=\frac{1}{5}m{v}_s{}^2=\frac{1}{5}\times m\times \frac{10}{7}gh=\frac{2}{7}mgh$

  $K{E}_{rotational/cylinder}=\frac{1}{4}m{v}_c{}^2=\frac{1}{4}\times m\times \frac{4}{3}gh=\frac{1}{3}mgh$

Therefore, the rotational energy of the cylinder is greater than that of the sphere

Conclusion

Therefore, (a) the sphere reaches the bottom first, (b) the sphere also has the greater speed at the bottom, (c) the total kinetic energy is the same for both at the bottom, and (d) the rotational kinetic energy of the cylinder is greater than that of the sphere.