Question

A thin uniform rod has a length of 0.430 m and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.32 rad/s and a moment of inertia about the axis of 3.20×10−3 kg⋅m2 . A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is 0.108 m/s . The bug can be treated as a point mass. Part A: what is the mass of the rod? Correct answer 5.19 x 10^-2 kg Part B: what is the mass of the bug?

Answer 1

a) Moment of inertia of a rod about an endpoint

I = 1/3 * m * L^2

m = 3 * I / L^2

m = 3 * 3.20 * 10^(-3) / 0.430^2

mass of rod m = 0.0519 kg = 5.19*10^(-2) kg

b) The angular momentum L is conserved:

The new angular velocity is

w = v/R = 0.108 m/s / 0.43 m = 0.251 rad/s

L1 = I * w1 = 3.20 * 10^(-3) * 0.32

L2 = I * w2 + I(bug) * w2

= 3.20 * 10^(-3) * 0.251 + m(bug) * 0.43^2 * 0.251

I(bug) = mL^2

L1 = L2

3.20 * 10^(-3) * 0.32 = 3.20 * 10^(-3) * 0.251 + m(bug) * 0.43^2 * 0.251

m(bug) = ( 3.20 * 10^(-3) * 0.32 - 3.20 * 10^(-3) * 0.251) / (0.43^2 * 0.251)

m(bug) = 4.757 * 10^(-3) kg