A thin uniform rod has a length of 0.480 m and is rotating in a circle on a fric

Question

A thin uniform rod has a length of 0.480 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 0.35 rad/s and a moment of inertia about the axis of 3.30×103 kgm2 . 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 0.110 m/s . The bug can be treated as a point mass.

Part A

What is the mass of the rod?

Express your answer with the appropriate units.

Part B

What is the mass of the bug?

Express your answer with the appropriate units.

Explanation / Answer

A)
moment of inertia of rod along its end,
Irod = M*L^2/3
3.3*10^-3 = M* (0.480)^2/3
M = 0.043 Kg
Answer: 0.043 kg

B)
use conservation of angular momentum
Initial momentum = final momentum
Irod*wi = (Irod + Ibug)*wf

wf = v/L = 0.110/0.480= 0.23 rad/s
I bug = m*L^2

Irod*wi = (Irod + Ibug)*wf
3.3*10^-3 * 0.35 = (3.3*10^-3 + m*L^2) * 0.23
3.3*10^-3 * 0.35 = (3.3*10^-3 + m*0.480^2) * 0.23
3.3*10^-3 + m*0.480^2 = 5.022*10^-3
m = 7.47*10^-3 kg
Answer: 7.47*10^-3 kg

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