A bike is turned upside down for repairs. The front wheel can be described as a

Question

A bike is turned upside down for repairs. The front wheel can be described as a hoop with radius R=0.6 m and mass M =1.0 kg and a small point mass m=0.005 kg attached to the perimeter to represent the valve for inflating the tire (see Fig. 1). We consider rotation of the wheel about its frictionless axle. Describe the two equilibrium positions for the wheel and identify the one of these that is a stable equilibrium. Calculate the period for small angle oscillations of the wheel about the stable equilibrium.

Explanation / Answer

a) The two equilibrium positions of the wheel are those in which the line joining the valve and the centre of the wheel coincide with the direction of gravity.

Out of these two, the position of stable equilibrium is that in which the valve is at the bottom. This is because any small rotational displacement of the wheel will cause the wheel to come to the same equilibrium position.

b) Restoring torque about the centre of the wheel, = -mgR(sin)

For small angle ,

= -mgR

=> = I = (MR2 + mR2) = -(mgR)

=> = -{mg/[(m + M)R]}

=> d2/dt2 = -{mg/[(m + M)R]}

=> d2/dt2 = -2

where = {mg/[(m + M)R]}1/2 = 2/T

So, time period of oscillations, T = [42(m + M)R/mg]1/2 = 42 * (1 + 0.005) * 0.6 / (0.005 * 9.81) = 22 s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.