Conservation of Angular Momentum of Bug Problem

  • Thread starter tylertwh
  • Start date
  • #1
22
0

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 = 0.12 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

This is the incorrect answer. I am not quite sure where I made my mistake?
All help would be appreciated!!
 

Answers and Replies

  • #2
50
0
Three problems that I can see.

The first is that the length of the bar needs to be in terms of meters. You've changed 120cm into 0.120 meters which is not right. 120cm=1.20m

The second is that the change is the fly's linear momentum transfers to the bars angular momentum. Find the fly's change in linear momentum (easy) and set it equal to the bar's angular momentum (slightly more work) and then solve for the angular speed of the bar.

The third is that the moment of inertia for a bar which rotates about its end is I=(1/3)ML^2. You forgot the 1/3

See if that helps
 
Last edited:
  • #3
22
0
ΔLinear Momentum (fly) = ΔAngular Momentum (bar)

mv(final) - mv(initial) = Iω(final) - Iω(initial)
= mr^2ω(final) - mr^2ω(initial)

0.013*0.025 - 0 = 0.065*1.2^2ω(final) - 0
3.25E-4 = 0.0936ω(final)
ω(final) = 0.00347

Does this look better?
 
  • #4
50
0
I apologize, i just made an edit on my last post. Look at what I wrote about the moment of inertia for the bar.

Also, it's not as straight forward as I thought. You need to figure out the torque that results as the fly jumps off of the rod. It's related to the force that the fly applies to the bar as it jumps off as well at the distance down the bar in which the fly jumped.

I will keep looking at this and get back to you although I'm sure somewhat will read this and correct me before I get another chance.
 
Last edited:
  • #5
22
0
So when I calculate the torque that the fly applies to the bar how would I integrate it into finding the angular velocity? Im just confused how the torque would be added to the angular velocity
 
  • #6
22
0
[itex]\tau[/itex] = rFsin(theta)

since here we know that theta = 90...

[itex]\tau[/itex] = rF

would F here be = ma(centripetal) = (mv^2)/r

**All for the bug**
 
  • #7
50
0
Try conservation of energy. KE of fly = KE of bar

If this doesn't work out I promise to stop trying to help
 
  • #8
22
0
Alright. No problem.

Useful Equations


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

how confident are you in this answer...
I am running out of "tries" in my online homework.
(no offense meant; I wouldn't have had any idea of other strategies w/o your help)
 
  • #9
50
0
I'm pretty darn confident this time. Conservation of energy never fails. The reason we can't use cons of momentum is because you'll need to know the torque applied by the fly which requires us to know the force the fly applied on the bar. And in order to figure out the force we would need to know the amount of time the fly applied the force which we don't have.

The alternative is to wait a little while and see if anyone corrects me or has a better idea.
 
  • #10
22
0
Well... unfortunately that answer was incorrect also.... I am not sure where to go with this, but I did repost this question as a new thread.
 
  • #11
50
0
I apologize, I'm obviously missing something that's leading me down the wrong path. Let me know how to properly do it when you figure it out.
 
  • #12
mukundpa
Homework Helper
524
3
I think you were correct for the the angular momentum of the bug about the axis of rotation after jump as mvL

The angular angular momentum of the rod after jump will be Iω = ML2/3*ω

Now apply law of conservation of angular momentum.
 

Related Threads on Conservation of Angular Momentum of Bug Problem

Replies
1
Views
7K
Replies
11
Views
4K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
2K
Replies
3
Views
1K
Replies
7
Views
4K
Replies
0
Views
6K
Replies
5
Views
666
  • Last Post
Replies
5
Views
2K
Top