Question

A cyclist turns her bicycle upside down to repair it. She then notices that the front wheel is executing a slow, small amplitude, back-and-forth rotational motion with period 12 s. Treating the wheel as a thin ring of mass 600 g and radius 30 cm, whose only irregularity is the tire valve stem, determine the mass of the valve stem.

Answer 1

The equation of motion you will have to use is

I d^2phi / dt^2 = T

where phi is the angle between the vertical and the line from the center to the valve (phi is zero when the valve is at the lowest point) , I is the moment of inertia of the wheel plus valve and T the torque. Denote the mass of the wheel with M and the mass of the valve with m. The radius of the wheel is R. Then the moment of inertia for rotation about the center of the wheel is

I = (M + m)R^2,

because all the mass is at distance R from the center of rotation.

The torque comes from the weight of the valve:

T = - m g R sin(phi)

So

(M+m)R^2 d^2phi/dt^2 = - m g R sin(phi)

As usual we consider small oscillations, so that sin(phi) can be well approximated with phi:

(M+m)R^2 d^2phi/dt^2 = - m g R phi
d^2phi /dt^2 + ( mg/((m+M)R)) phi = 0

From this we read off:

Omega^2 = mg/((m+M)R))

and using Omega^2 = (2 pi /T)^2

4 pi^2 / T^2 = mg/((m+M)R)

4 pi^2 / T^2 = g/((1+M/m)R)

1+M/m= g T^2 /( 4 pi^2 R)

M/m = g T^2 /( 4 pi^2 R) - 1

m = M/( g T^2 /( 4 pi^2 R) - 1)

Using the data given we therefore obtain

m = 600 gram /(9.81m/s^2*(12s)^2/( 4*3.1415^2*0.30m) -1)
m= 4.32 gram