Two spheres of unequal masses but of the same radii are released from the top of a smooth inclined plane. They roll down the plane without slipping. Which one will reach the bottom first?

Options

A. Both will reach the bottom at the the same time
B. Heavier sphere
C. Lighter sphere
D. None of the above

Answer

A

Explanation

The accelertion of the body rolling down a smooth inclined plane is given by
`a=(g sin theta)/(1+(K^(2))/(r^(2)))`
and its velocity when it reaches the bottom is
`v=sqrt((2gh)/((1+(K^(2))/(r^(2)))))`
and the time to reach the bottom is given by
`therefore t=(v)/(a)" "(because u=0)`
`therefore t=sqrt((2gh)/((1+(K^(2))/(r^(2))))xx((1+(K^(2))/(r^(2)))^(2))/(g^(2)sin^(2)theta))`
`therefore t=(1)/(sin theta)sqrt((2h(1+(K^(2))/(r^(2))))/(g))`
Thus the time to reach the bottom is independent of the mass of the sphere.
`therefore` Both spheres will reach the bottom at the same time.