Question

A body slides down a smooth inclined plane having angle θ and reaches the bottom with velocity v. If a body is a sphere, then its linear velocity at the bottom of the plane is

Options

• `sqrt(9/7)`v

• `sqrt(5/7)`v

• `sqrt(2/7)`v

• `sqrt(3/7)`v

Answer 1

`sqrt(5/7)`v

Explanation:

The linear velocity of the body, v = `sqrt(2"gh")`

The velocity of the sphere about its centre,

`"v"_"CM" = sqrt((2"gh")/(1+ "K"^2/"R"^2)) = "v"/sqrt(1 + "K"^2/"R"^2)`   ...(i)

For uniform solid sphere,

`"K"^2/"R"^2 = 2/5`

Substituting value in Eq. (i), we get

`"v"_"CM" = "v"/sqrt(1 + (2/5)) = sqrt(5/7)`v