Question

A metal disc of radius R and mass M freely rolls down from the top of an inclined plane of height h without slipping. The speed of its centre of mass on reaching the bottom of the inclined plane is

Answer 1

Answer:

Energy should be conserved

Potential energy = Rotational K.E + Translational K.E

$$mgh = \dfrac{1}{2}Iw^2 + \dfrac{1}{2} mv^2 $$

$$mgh = \dfrac{1}{4} mr^2 w^2 + \dfrac{1}{2} mv^2 $$

vcm=rw

$$gh = \dfrac{3}{4} v^2  $$

$$v = \sqrt{ \dfrac{4}{3} gh } $$

Answer 2

Energy should be conserved

Potential energy = Rotational K.E + Translational K.E

$mgh = \dfrac{1}{2}Iw^2 + \dfrac{1}{2} mv^2$

$mgh = \dfrac{1}{4} mr^2 w^2 + \dfrac{1}{2} mv^2$

vcm=rw

$gh = \dfrac{3}{4} v^2$

$v = \sqrt{ \dfrac{4}{3} gh }$