Question

A wheel 1.55 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 3.65 rad/s². The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3

Answer 1

the radius of the wheel is 1m

angular velocity = initial angular velocity + A t where A is angular accel and t is time

ang vel = 0 + 3.75rad/s/s x 2s = 7.5 rad/s

b) tangential vel = r w = 0.775m x 7.5 rad/s = 5.8125m/s

c) total accel = Sqrt[a tan^2 + a radial^2]

a tan = 3.75rad/s/s, a radial = v^2/r = (5.812m/s)^2/0.775 m =43.593m/s/s

d) total angular displacement of the point P = w0 t + 1/2 A t^2 = 0 + 1/2 (3.75rad/s/s)(2s)^2 = 7.5 rad

the initial position, 57.3 deg = 1 rad, so final position = 8.5 rad = 127 deg with respect to the horizontal