Question

A uniform disc of mass M=2.0 kg and radius R=0.45 m rolls without slipping down an inclined plane of length L=40 m and slope of 30°. The disk starts from rest at the top of the incline. Find the angular velocity at the bottom of the incline.

Answer 1

here,

mass of disc , m = 2 kg

radius , r = 0.45 m

L = 40 m

theta = 30 degree

let the angular velocity at the bottom of the incline be w

using conservation of energy

(0.5 * I * w^2 + 0.5 * m * v^2) - (0.5 * I * w0^2 + 0.5 * m * u^2) = m * g * (L * sin(theta))

(0.5 * 0.5 * m * r^2 * (v/r)^2 + 0.5 * m * v^2) - (0.5 * I * 0^2 + 0.5 * m * 0^2) = 2 * 9.81 * (40 * sin(30))

0.75 * v^2 = 9.81 * 40 * sin(30)

solving for v

v = 16.2 m/s

the angular velocity of disk at the bottom , w = v /r

w = 16.2 /0.45 rad/s = 35.95 rad/s