# Bug Jumps off a rod; what is the angular velocity of the rod

## Homework Statement

A small 13.0g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 65.0g and is 120cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 25.0cm/s relative to the table.

mass bug = 0.013 kg
mass bar = 0.065 kg
distance from axis of rotation = 1.2 m
velocity bug = 0.025 m/s

Find
What is the angular speed of the bar just after the frisky insect leaps?

## Homework Equations

L = Iω
I(bug) = Mr^2
ω(bug) = v/r
L(bug) = Iω = Mr^2ω = Mrv
I(bar) = Mr^2
ω(bar) = v/r
L(bar) = Iω = Mr^2ω

## The Attempt at a Solution

Conservation of Angular Momentum

L(bug) = L(bar)
M*r*v = M*r^2*ω
0.013*0.12*0.025 = 0.065*0.12^2*ω
3.9*10^5 = 9.36*10^4ω
ω = 0.204

Conservation of Kinetic Energy

KE = 1/2Iω^2

KE(bug) = 1/2Iω^2
KE(bug) = 1/2(mr^2)ω^2
KE(bug) = 1/2(mr^2)(v/r)^2
KE(bug) = 1/2mv^2

KE(bar) = 1/2Iω^2
KE(bar) = 1/2(1/3mr^2)ω^2
KE(bar) = 1/6mr^2ω^2

KE(bug) = KE(bar)
1/2mv^2 = 1/6mr^2ω^2
4.0625E-6 = 0.0156ω^2
ω^2 = 2.6042E-4
ω = 0.01614

## Answers and Replies

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mfb
Mentor
Kinetic energy is not conserved, the bug has to use some source of energy to jump. There is no particular reason to expect the same kinetic energy for both bug and bar.
Conservation of angular momentum works - but you have to consider that the mass of the bar is not at its end only, it is distributed over the whole bar.

Do you know the right answer? If you divide your value (for conservation of angular momentum) by the right answer, you should note some relation between the two.

Also, you said the moment of inertia of the bar is mr^2 - I don't think that's right. If you integrate r^2*dm for a uniform thin bar, you get something slightly different.

Edit: think mfb and I are saying the same thing: re-check your moment of inertia for the bar.

Last edited:
So if I change the moment of inertia of the bar to 1/3mr^3 then I should get the correct answer?

1/3mr2 should do it :)

Why is it r2 and not r3?

because I thought you would take the integral of mr2 and in order to get it outside of the $\Sigma$ you had to take the anti-derivative which would be 1/3mr3?

L = Iω
I(bug) = Mr^2
ω(bug) = v/r
L(bug) = Iω = Mr^2ω = Mrv
I(bar) = 1/3*Mr^2
ω(bar) = v/r
L(bar) = Iω = 1/3Mr^2ω

L(bug) = L(bar)
M*r*v = (1/3)*M*r^2*ω
0.013*0.12*0.025 = 0.02167*0.12^2*ω
3.9*10^-5 = 3.12*10^-4ω
ω = 0.125

First, r2 is dimensionally correct, but the reason also comes from

∫r2dm.

dm = (m/L)*dx, and r = x, so we have

(m/L) * ∫x2dx. Integrate from 0 to L.

What does that work out to?

First, r2 is dimensionally correct, but the reason also comes from

∫r2dm.

dm = (m/L)*dx, and r = x, so we have

(m/L) * ∫x2dx. Integrate from 0 to L.

What does that work out to?
(m/L) * (1/3)L3 - (1/3)03

which when you simplify is M*(1/3)L2

Thank you so much for your help. I will post if the answer is correct or not shortly

(m/L) * (1/3)x3

Use the limit - evaluate that between x=0 and x=L. Notice one of the L's cancels.

Yeah I noticed that when I made the reply. I edited my previous reply! Thank you so much!

The answer was correct!

Thank you very much!