Question

With axle and spokes of negligible mass and a thin rim, a certain bicycle wheel has a radius of 0.350 m and weighs 33.0 N; it can turn on its axle with negligible friction. A man holds the wheel above his head with the axle vertical while he stands on a turntable that is free to rotate without friction; the wheel rotates clockwise, as seen from above, with an angular speed of 66.0 rad/s, and the turntable is initially at rest. The rotational inertia of wheel + man + turntable about the common axis of rotation is 2.00 kg·m2. The man's free hand suddenly stops the rotation of the wheel (relative to the turntable). (a) Determine the resulting angular speed of the system. (b) Is the direction of the rotation clockwise or counterclockwise?

Answer 1

mass of wheel m = weight/g = 33N/9.8m/s²=3.37 kg

radius of wheel r = 0.350 m

angular speed ω ₁ = 66rad/s

moment of inertia OF WHEEL I₁= mr²

initial angular momentum L₁ = I₁ω₁ = mr²ω₁ = 3.37*0.1225*66kgm²/sec

                                                                          = 27.225 kgm²/sec--1

total moment of inertia of the system (wheel + man +turntable) I₂= 2.00 kg·m².

Suppose the final angular speed be ω ₂

final angular momentum L2 = 2.00 * ω ₂kg·m²/sec------2

From law of conservation of angular momentum:

1 =L2

27.225 kgm²/sec = 2.00 * ω₂ kg·m²/s

SOLVE FOR ω ₂ =13.6 rad/s

b)to conserve angular momentum that should be opposite indirection of rotation of wheel

that is counter clock wise because the wheel rotates clockwise