# Simple bug on uniform rod rotational momentum problem

1. Sep 25, 2011

### area5x1

1. The problem statement, all variables and given/known data

A small 10.0-g 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 60.0g and is 90cm 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.

2. Relevant equations

L = Iw
P = mv

3. The attempt at a solution

I calculated the linear momentum of the bug: p=0.01(.25) = 0.0025

I then set that equal to the momentum of the rod: 0.0025 = Iw. The moment of inertia for the rod is I=(1/3)MR^2. Therefore I = (1/3)(0.06)(0.9)^2 = 0.0162.

0.0025 = (0.0162)w. Solve for w. I get w = 0.154 rad/s. I have tried both positive and negative values for w and neither are correct. What am I doing wrong? This is for Mastering Physics, so the values are a little bit different from the textbook. However, when I plugged in textbook values, I get the same answer as the answer in the back of the book, so my method doesn't seem to be wrong...

2. Sep 25, 2011

### dynamicsolo

You need the relationship for the angular momentum of an object of mass m and velocity v moving on a straight line at a perpendicular distance D from a reference point. Since L = r x p , the magnitude of this angular momentum is L = rmv sin(theta) . We can regroup this into L = mv · [ r sin(theta) ] . If we measure the moment arm r from the reference point for rotation to the object anywhere along its straight line of travel, and measure the angle theta between that moment arm's vector and the velocity vector, we find that it produces a constant [ r sin(theta) ] = D , with D being the perpendicular distance from the reference point to the line of travel.

Since the bug leapt from on a line of travel perpendicular to the bar, the perpendicular distance from the rotation axis to the line of travel is just the length of the bar, or D = R . So the angular momentum of the bug after the leap is L = mvR . The bug and bar were initially at rest, so the bar will "recoil" with an angular momentum in the opposite direction but also with magnitude L = mvR .

So you want to set mvR = (1/3)MR2w for the bar.