Conservationof energy and angular speed

[CornMuffin]

A small 13.0-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 70.0 g and is 90 cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 25.0 cm/s relative to the table.

Homework Statement

What is the angular speed of the bar just after the frisky insect leaps?
mass of bug = .013kg
mass of bar = .070kg
length of bar = 0.9m
final velocity of bug = 0.25 m/s
initial velocity of bug = 0
initial velocity of bar = 0

Homework Equations

I = (1/3)ML^2
K(bug) = (1/2)mV^2
K(bar) = (1/2)Iw
Sum of the energy before = sum of the energy after

The Attempt at a Solution

0 = K(bug) + K(bar)
0 = (1/2)mV^2 + (1/2)Iw
0 = (1/2)mV^2 + (1/2)(1/3)ML^2w
w = -[mV^2]/[(1/3)ML^2]

but that doesn't come out with the correct answer

where w is the lowercase omega standing for angular velocity

 

[hage567]

Just quickly looking at this, I notice you left the square off the omega in your equation for rotational kinetic energy. It should be K(bar) = 0.5Iw^2.

 

[CornMuffin]

Ya, i noticed that a little bit ago, and when I put in the square... This is from a problem in the book with different values, but using the values in the book, my answer still does not agree with the answer in the book. If i did everything right, maybe the answer in the book is wrong...it could happen lol.

 

[CornMuffin]

Well, the final answer turned out to be 0.155 rad/s, but I'm not sure why

 

[hage567]

Well, the final answer turned out to be 0.155 rad/s, but I'm not sure why

So this is the book's answer?

 

[CornMuffin]

So this is the book's answer?

yes, using these values, that is the answer

 

[hage567]

Was the book using conservation of energy in the example? I would use conservation of angular momentum about the pivot point for this problem.