Answer
(a) The spheres reach the bottom at the same time.
(b) The translational speed will be the same for each sphere. The rotational speed will be smaller for the sphere with the larger radius.
(c) The sphere with the larger mass will have a larger amount of total kinetic energy.
Work Step by Step
As a sphere rolls down an incline, the initial potential energy is converted into translational and rotational kinetic energy. We can use conservation of energy to solve this question.
$KE_t + KE_r = PE$
$\frac{1}{2}mv^2+ \frac{1}{2}I\omega^2 = mgh$
$\frac{1}{2}mv^2+ \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$
$\frac{1}{2}mv^2+ \frac{1}{5}mv^2 = mgh$
$\frac{7}{10}v^2 = gh$
$v = \sqrt{\frac{10gh}{7}}$
(a) At any point along the incline, the two spheres will have the same speed, so they must reach the bottom at the same time.
(b) At the bottom of the incline, the translational speed of each sphere is the same. The translational speed is $v = \sqrt{\frac{10gh}{7}}$.
However, the rotational speed will be different.
$\omega = \frac{v}{r}$, so the sphere with a larger radius will have a slower rotational speed.
(c) At the bottom, the total kinetic energy for each sphere is $KE = \frac{7}{10}mv^2$. The speed $v$ is the same for each sphere, so the sphere with the larger mass will have a larger amount of total kinetic energy.
