Question

An object of mass m = 3.05kg rolls off of an incline plane from a height of 7.55m, where the angle of the incline measured from the horizontal is 37.3o. If the object is a solid uniform cylinder of radius r = 1.25m, then at the bottom of the ramp, what is the (a) rotational kinetic energy and (b) the translational kinetic energy. If the object is a hoop of radius, r = 1.05m, then at the bottom of the ramp, what is the (c) rotational kinetic energy and (d) the translational kinetic energy?

Answer 1

Part a)

1/2 m v^2 + 1/2 I w^2 = m g h

1/2 m (wR)^2 + 1/2 (1/2 m R^2) w^2 = m g h

1/2 (wR)^2 + 1/4 (R^2) w^2 = g h

3/4 (wR)^2 = g h

w = √(4gh/3)/R = √(4*9.81*7.55/3)/1.25 = 7.95 rad/s

I = 1/2 m R^2 = 0.5 * 3.05 * 1.25 * 1.25 = 2.3828 Kg.m²

K = 1/2 I w^2 = 0.5 * 2.3828 * 7.95*7.95

Rotational Kinetic Energy of a solid uniform cylinder is, K = 75.30 J

Part b)

w = √(4gh/3)/R = √(4*9.81*7.55/3)/1.25 = 7.95 rad/s

v = wR = 7.95*1.25 = 9.9375 m/s

Translational Kinetic Energy, K = 1/2 * m*v²

K = 0.5*3.05*9.9375²

Translational Kinetic Energy of a solid uniform cylinder is, K = 150.60 J

Part c)

1/2 m v^2 + 1/2 I w^2 = m g h

1/2 m (wR)^2 + 1/2 (m R^2) w^2 = m g h

1/2 (wR)^2 + 1/2 (R^2) w^2 = g h

(wR)^2 = g h

w = √(gh)/R = √(9.81*7.55)/1.05 = 8.1963 rad/s

I = m R^2 = 3.05 * 1.05 * 1.05 = 3.3626 Kg.m²

K = 1/2 I w^2 = 0.5 * 3.3626 * 8.1963*8.1963

Rotational Kinetic Energy of a hoop is, K =  112.95 J

Part d)

w = √(gh)/R = √(9.81*7.55)/1.05 = 8.1963 rad/s

v = wR = 8.1963*1.05

v = 8.61 m/s

Translational Kinetic Energy, K = 1/2 * m*v²

K = 0.5*3.05*8.61²

Translational Kinetic Energy of a hoop is, K = 112.95 J