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For a hollow cylinder and a solid cylinder rolling without slipping on an inclined plane, then which of these reaches earlier on the ground :
A
Solid cylinder
B
Hollow cylinder
C
Both simultaneously
D
Can't say anything
Detailed Solution
For a body rolling down an incline: $a = \dfrac{g\sin\theta}{1 + K^2/R^2}$
Solid cylinder: $\dfrac{K^2}{R^2} = \dfrac{1}{2}$, so $a = \dfrac{2}{3}g\sin\theta$
Hollow cylinder: $\dfrac{K^2}{R^2} = 1$, so $a = \dfrac{1}{2}g\sin\theta$
The solid cylinder has the larger acceleration (less of its energy goes into rotation), so it reaches the ground first.
So the solid cylinder reaches earlier.
Solid cylinder: $\dfrac{K^2}{R^2} = \dfrac{1}{2}$, so $a = \dfrac{2}{3}g\sin\theta$
Hollow cylinder: $\dfrac{K^2}{R^2} = 1$, so $a = \dfrac{1}{2}g\sin\theta$
The solid cylinder has the larger acceleration (less of its energy goes into rotation), so it reaches the ground first.
So the solid cylinder reaches earlier.