A bicycle wheel of radius 0.36 m and mass 3.2 kg is set spinning at 4.00 rev/s.A very light bolt is attached to extend the axle in length,and a string is attached to the axle at a distance of 0.10 m from the wheel.Initially the axle of the spinning wheel is horizontal,and the wheel is suspended only from the string.We can ignore the mass of the axle and spokes.At what rate will the wheel process about the vertical?
A)2.9 rpm
B)1.9 rpm
C)18 rpm
D)0.30 rpm
E)0.77 rpm

Question

Quizplus · Accepted Answer

The correct answer is A.Explanation:The initial angular momentum of the wheel is:$$L_i = I_i \omega_i = \frac{1}{2}mr^2 \omega_i$$where:* $I_i$ is the initial moment of inertia of the wheel* $m$ is the mass of the wheel* $r$ is the radius of the wheel* $\omega_i$ is the initial angular velocity of the wheelWhen the bolt is attached, the moment of inertia of the wheel increases to:$$I_f = I_i + mr_b^2$$where:* $r_b$ is the distance from the axle to the boltThe final angular velocity of the wheel is:$$\omega_f = \frac{L_i}{I_f} = \frac{\frac{1}{2}mr^2 \omega_i}{\frac{1}{2}mr^2 + mr_b^2}$$Substituting the given values, we get:$$\omega_f = \frac{\frac{1}{2}(3.2 	ext{ kg})(0.36 	ext{ m})^2 (4.00 	ext{ rev/s})}{\frac{1}{2}(3.2 	ext{ kg})(0.36 	ext{ m})^2 + (3.2 	ext{ kg})(0.10 	ext{ m})^2} = 2.9 	ext{ rev/s}$$Therefore, the wheel will process about the vertical at a rate of 2.9 rpm.

A Bicycle Wheel of Radius 0