Question

[Use g = 10 m/s^2] A non-uniform wheel of mass 5 kg and moment of inertia 1/3 mR^2 is set on an incline whose height is h = 4 meters and length is L = 20 meters. The wheel is released from rest at the top of the incline and rolls without slipping to the bottom. What is the wheel's translational kinetic energy at the bottom of the incline? What is the wheel's rotational kinetic energy at the bottom of the incline? What is the wheel's constant acceleration as it rolls down the incline? What is the frictional force acting on the wheel as it rolls down the incline?

Answer 1

M = 5kg, I = 1/2 mR^2 h = 4m, L = 20M (a) using conservation of energy Potential energy at Rotation Kinetic energy top = + transitional energy mgh = 1/2 I w^2 + 1/2 mv^2 = 1/2 (1/2 m R^2) (V/R)^2 + 1/2 MV^2 = 1/2 MV^2 [1/2 + 1] 1/2 mv^2 = Translational energy Translational energy = 2/3 mgh = 2/3 times 5 times 9.81 times 4 = 130.8 J (b) Rotational KE = mgh - 1/2mv^2 = (5 times 9.8 times 4 - 130.8)J = 65.4J (c) According to fig ma = mg sin theta -f According to torque 7R = I_a/r_v of = I_a/R^2 of = mgsin theta -ma 1/2 mR^2 a/R^2 = mg sin theta - ma 3ma/2 = mg sin theta a = 2/3 g sin theta = 2/3 g (h/l) a = 1.31 m/s^2 (d) frictional force f = mgsin theta - ma 5 times 9..81 times 4/20 - 5 times 1.31 f = 3.27N