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A solid cylinder and a hollow cylinder, both of the same mass and same external diameter, are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?
A
Both together only when angle of inclination of plane is $45^\circ$
B
Both together
C
Hollow cylinder
D
Solid cylinder
Detailed Solution
Acceleration of a body rolling without slipping down an incline: $a = \frac{g\sin\theta}{1 + \frac{k^2}{R^2}}$, where k is the radius of gyration.
Time to cover a length l of the incline from rest: $t = \sqrt{\frac{2l}{a}} = \sqrt{\frac{2l\left(1 + \frac{k^2}{R^2}\right)}{g\sin\theta}}$
Solid cylinder: $I = \frac{1}{2}MR^2$, so $\frac{k^2}{R^2} = \frac{1}{2}$
Hollow cylinder: its mass is farther from the axis, so $\frac{k^2}{R^2}$ is larger (it equals 1 for a thin shell).
A smaller $\frac{k^2}{R^2}$ gives a larger acceleration and a shorter time, independent of mass and of the angle of inclination.
Hence the solid cylinder reaches the bottom first.
Time to cover a length l of the incline from rest: $t = \sqrt{\frac{2l}{a}} = \sqrt{\frac{2l\left(1 + \frac{k^2}{R^2}\right)}{g\sin\theta}}$
Solid cylinder: $I = \frac{1}{2}MR^2$, so $\frac{k^2}{R^2} = \frac{1}{2}$
Hollow cylinder: its mass is farther from the axis, so $\frac{k^2}{R^2}$ is larger (it equals 1 for a thin shell).
A smaller $\frac{k^2}{R^2}$ gives a larger acceleration and a shorter time, independent of mass and of the angle of inclination.
Hence the solid cylinder reaches the bottom first.