How does a jumping bug affect the rotational dynamics of a uniform bar?

[eeriana]

Homework Statement

A bug of mass 10.0 g 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 50.0g and is 100 cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a 20.0 cm/s relative to the table. a) What is the angular speed of the bar just after the bug leaps? b) What is the total kinetic energy of the system just after the bug leaps? c) Where does this energy come from?

Homework Equations

Ke= 1/2 MVcm^2

$$\omega$$=V/r

The Attempt at a Solution

First I want to say thanks, you all have been a big help to me with the questions I have posed in the past... I don't always say it but thank you.

My problem is trying to set it up, I tried to draw the picture and can't visualize the orientation of the system. I can't seem to figure out how the bar rotates if it is horizontal then how can the bug jump off in the horizontal direction perpendicular to the bar. Any thoughts to help me put this in perspective?

 

[catkin]

Think of it as a clock lying face up on a horizontal surface with the minute hand (the bar)pointing to 12. The bug, sat on the end of the hand furthest from the pivot, jumps horizontally, say toward 11.

There are many "perpendicular to the bar"s, a whole 360 degrees of them, but only two are horizontal.

 

[eeriana]

So when I draw it, I should draw it from a top view? Would that let me see what's going on from a better perspective?

 

[catkin]

Sure

 

[eeriana]

Thank you so much I got the angular velocity... that makes sense now.

One more thing... I can't seem to get the KE in the system after the bug jumped. I know the answer in the book, I am using: KE=1/2mv^2 + 1/2I$$\omega$$^2. I thought I was on the right track and then got stumped. Am I not looking at this right?

Thanks again...

 

[catkin]

I don't (yet) do rotational dynamics so will leave that for someone else to answer.

Sometimes nobody else will look in on a thread where someone is already helping so you might want to start a new thread with your remaining question.

One last thing ... what sort of bug weighs 10g?!

 

[eeriana]

HAHA I don't know. I don't want to know. heh. But thank you for your help!

 

[Thomas_]

I have the same problem. I know that the bug produces a torque (I don't know what torque though), and therefore the Angular Momentum is not conserved. However, I can't figure out how this helps me. Does anyone have a hint for this problem?

 

[catkin]

Be impulsive!

What impules is required to accelerate the bug? There must be an equal and opposite impulse on the bar.

 

[Shooting Star]

Thank you so much I got the angular velocity... that makes sense now.

One more thing... I can't seem to get the KE in the system after the bug jumped. I know the answer in the book, I am using: KE=1/2mv^2 + 1/2I$$\omega$$^2. I thought I was on the right track and then got stumped. Am I not looking at this right?

Thanks again...

If you've already got the angular velo w, then simply calculate Iw^2/2 for the rotational KE of the rod, and mv^2/2 for the KE of the bug. Isn't that giving the right answer?

The energy comes from the muscular energy of the bug.