Question

A solid cylinder of radius r and mass M rolls down an inclined plane of height h. When it reaches the bottom of the plane, then its rotational kinetic energy is ____________.

(g = acceleration due to gravity)

Options

• `("Mgh")/4`

• `("Mgh")/2`

• Mgh

• `("Mgh")/3`

Answer 1

A solid cylinder of radius r and mass M rolls down an inclined plane of height h. When it reaches the bottom of the plane, then its rotational kinetic energy is `underline(("Mgh")/3)`.

Explanation:

When the solid cylinder reaches the bottom of the plane, its rotational kinetic energy is given by

K_r = `1/2"l"ω^2`

As, l = Mk² and ω = `("v"_"CM")/"R"`

K_r = `1/2"Mk"^2 xx (("v"_"CM")/"R"^2)^2`

= `1/2 "Mv"_"CM"^2 ("k"^2/"R"^2)`

At bottom, velooity, v_CM = `sqrt((2"gh")/(1 + "k"^2/"R"^2))`

Also, for solid cylinder, `"k"^2/"R"^2 = 1/2`

∴ K_r = `1/2 "M" [(2"gh")/((1 + 1/2))] (1/2) = (4  "Mgh")/(4 xx 3) = "Mgh"/3`