Precession: 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.The moment of inertia of the wheel about the vertical axis is:$$I = mr^2 + ma^2 = (3.2 	ext{ kg})(0.36 	ext{ m})^2 + (3.2 	ext{ kg})(0.10 	ext{ m})^2 = 0.461 	ext{ kg m}^2$$The angular momentum of the wheel is:$$L = I\omega = (0.461 	ext{ kg m}^2)(4.00 	ext{ rev/s})(2\pi 	ext{ rad/rev}) = 18.4 	ext{ kg m}^2/	ext{s}$$The torque due to the string is:$$	au = mgd = (3.2 	ext{ kg})(9.81 	ext{ m/s}^2)(0.10 	ext{ m}) = 0.314 	ext{ N m}$$The rate of precession is:$$\omega_p = \frac{	au}{I} = \frac{0.314 	ext{ N m}}{0.461 	ext{ kg m}^2} = 0.681 	ext{ rad/s} = 2.9 	ext{ rpm}$$

Precession: a Bicycle Wheel of Radius 0