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

In summary, the conversation discussed the scenario of a small bug jumping off a uniform bar that is initially at rest on a smooth horizontal table. The bug has a mass of 13.0g and jumps off with a speed of 25.0cm/s relative to the table. The bar has a mass of 65.0g and is 120cm in length. The task was to find the angular speed of the bar just after the bug jumps off. The conversation involved using the equations for conservation of angular momentum and kinetic energy, and considering the distribution of mass along the bar to determine the moment of inertia. The correct answer was found to be 0.125 rad/s.
  • #1
tylertwh
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 = 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

So far none are correct. Please Help!
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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:
  • #4
So if I change the moment of inertia of the bar to 1/3mr^3 then I should get the correct answer?
 
  • #5
1/3mr2 should do it :)
 
  • #6
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 [itex]\Sigma[/itex] you had to take the anti-derivative which would be 1/3mr3?
 
  • #7
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
 
  • #8
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?
 
  • #9
Ampere said:
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
 
  • #10
(m/L) * (1/3)x3

Use the limit - evaluate that between x=0 and x=L. Notice one of the L's cancels.
 
  • #11
Yeah I noticed that when I made the reply. I edited my previous reply! Thank you so much!
 
  • #12
The answer was correct!

Thank you very much!
 

1. What is angular velocity?

Angular velocity is a measure of how fast an object is rotating around a fixed axis. It is usually measured in radians per second or degrees per second.

2. How is angular velocity related to the rod?

In this scenario, angular velocity refers to the rate at which the rod is rotating as the bug jumps off of it. It is determined by the speed and direction of the bug's jump and the length of the rod.

3. Can the angular velocity of the rod change?

Yes, the angular velocity of the rod can change if the bug jumps off at a different speed or in a different direction. It can also change if the length of the rod is adjusted.

4. How is angular velocity calculated?

Angular velocity can be calculated by dividing the change in angle (in radians or degrees) by the change in time. It can also be calculated using the formula ω = v/r, where ω is angular velocity, v is linear velocity, and r is the distance from the axis of rotation.

5. What factors affect the angular velocity of the rod?

The angular velocity of the rod can be affected by the speed and direction of the bug's jump, the length and mass of the rod, and any external forces acting on the rod (such as friction or gravity).

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
8K
  • Introductory Physics Homework Help
Replies
10
Views
803
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
903
  • Introductory Physics Homework Help
2
Replies
62
Views
9K
  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top