# A cannonball and a marble roll from rest down an incline. Which gets to the bottom first?

askIITians Faculty 396 Points
9 years ago
Both, the cannonball and the marble rolls from rest down the incline plane in the same time. It is obvious that the radius of the marble will be smaller than the radius of the cannonball, therefore the marble will have smaller rotational inertia than the cannonball and will make more revolutions per second.
The distance travelled by the cannonball in one revolution is much greater than the distance travelled by the marble. Therefore the motion of the balls is such that the larger distance travelled by the cannonball in each revolution is compensated by the large number of revolutions made by the marble in the same time.
Therefore the cannonball and the marble reach the bottom in the same time.
S S Mathur
19 Points
7 years ago
In the case of any ball rolling down an inclined place without slipping: the potential energy mgh at the top of the incline is equal to the KE at the bottom. The KE = translational KE ($\frac{1}{2}mv^{2}$) + rotational KE ($\frac{1}{2}I\omega ^{2}$).  For a sphere, $I=\frac{2}{5}mr^{2}$ and $\omega =\frac{v}{r}$. Therefore at the bottom of the incline velocity $v=\sqrt{\frac{10}{7}gh}$. Since $v^{2}=2as$ (s being the length of the inclined plane), and $\frac{h}{s}=sin\theta$ (the angle of the incline), acceleration $a=\frac{5gsin\theta }{7}$. Both velocity at the bottom and acceleration are independent of the mass or radius of the ball, and depend only on the length of the inclined plane and its angle. Therefore the time taken by both of them to reach the bottom is the same.