Question

A thin uniform rod has a length of 0.400 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 0.34 rad/s and a moment of inertia about the axis of 2.50×10−3 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 0.112 m/s . The bug can be treated as a point mass. Part APart complete What is the mass of the rod? Express your answer with the appropriate units. mrod = 4.69×10−2 kg Previous Answers Correct Part B What is the mass of the bug? Express your answer with the appropriate units. mbugm b u g =

Answer 1

Part A.

Moment of inertia of a rod about one endpoint is given by:

I = (1/3)*m*L^2

m = 3I/L^2

m = 3*2.50*10^-3/0.400^2

m = 0.0469 kg = 4.69*10^-2 kg

Part B.

Since there is no external force applied on bug + rod system, So net angular momentum will remain conserved, So

Li = Lf

Ii*wi = If*wf

Ii = 2.50*10^-3 kg.m^2, wi = 0.34 rad/sec

If = Ii + m_bug*r^2 = 2.50*10^-3 + m_bug*0.400^2

wf = Vf/R = 0.112/0.400 = 0.28 rad/sec

So,

2.50*10^-3*0.34 = (2.50*10^-3 + m_bug*0.400^2)*0.28

m_bug*0.400^2 = 2.50*10^-3*0.34/0.28 - 2.50*10^-3

m_bug = 5.357*10^-4/0.400^2

m_bug = 3.35*10^-3 kg = mass of bug

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