Angular Momentum and torque problem

  • Thread starter waterfaire
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  • #1
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The Problem:
Two identical spheres, each of mass M and negligible radius, are fastened to opposite ends of a rod of negligible mass and lenth 2l. This system is initially at rest with the rod horizontal, and is free to rotate about a frictionless horizontal axis through the center of the rod and perpendicular to the plan of the page. A bug, of mass 3M, lands gently on the sphere on the left, Assume that the size of the bug is small compared to the length of the rod. Express your answesr to all parts of the question in terms of M, L, and physical constants.

A) Determine the Torque about the axis immediately after the bug lands on the sphere

B) Determine the angular acceleration of the rod-sphere-bug system immediately after the bug lands

I got the Torque to be 3Mgl ... which is the right answer, but I dont know where to start for Part B



There is another part to this question...
The rod-sphere-bug system swings about the axis. at the instant that the rod is vertical, determine:
C) the angular speed of the bug
D) the angular momentum of the system
E) the magnitude and direction of the force that must be exerted on the bug by the sphere to keep the bug from being thrown off the sphere.

I have n o clue where to start...

Please help
 

Answers and Replies

  • #2
Doc Al
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I got the Torque to be 3Mgl ... which is the right answer, but I dont know where to start for Part B
Apply Newton's 2nd law for rotation. (What's the rotational inertia of the bug+spheres system?)

For the other parts: As the system rotates, is anything conserved?
 
  • #3
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As the system rotates...Kinetic energy is conserved?
 
  • #4
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Def. of Torque is...r F

or

I * alpha
 
  • #5
Doc Al
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As the system rotates...Kinetic energy is conserved?
If KE by itself was conserved, the system couldn't start moving, since it starts from rest. But total mechanical energy (KE + PE) will be conserved, since the axis is frictionless.
 
  • #6
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Ohh...So work is conserved then.
 
  • #7
Doc Al
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Ohh...So work is conserved then.
Don't know what you mean by that.
 
  • #8
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Well. Doesn't W = change in KE + change in PE?
 
  • #9
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Apply Newton's 2nd law for rotation. (What's the rotational inertia of the bug+spheres system?)

How do you find the I if the radius of the sphere is negligible?

for Part B.
I have T= I * alpha, and I substituted the answer i got for Part A (3Mgl) for T.

3Mgl = I alpha
 
  • #10
Doc Al
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Well. Doesn't W = change in KE + change in PE?
Sure. And if you track the effect of gravity via gravitational PE, there are no forces doing any work on the system so mechanical energy is conserved.

How do you find the I if the radius of the sphere is negligible?
I have no idea why they bothered to specify them as spheres if their radii were negligible. Treat them as point masses. (What's the rotational inertia of a point mass about some axis?)

for Part B.
I have T= I * alpha, and I substituted the answer i got for Part A (3Mgl) for T.

3Mgl = I alpha
Good. Now find I and you're on your way.
 
  • #11
Sure. And if you track the effect of gravity via gravitational PE, there are no forces doing any work on the system so mechanical energy is conserved.


I have no idea why they bothered to specify them as spheres if their radii were negligible. Treat them as point masses. (What's the rotational inertia of a point mass about some axis?)


Good. Now find I and you're on your way.
The moment of inertia for a thin, uniform rod of length L and mass m with its axis at its center is m*L^2/12. Does m = 2M, the mass of just the spheres, or m = 5M, the mass of the whole rod-spheres-bug system?
 
  • #12
Doc Al
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The moment of inertia for a thin, uniform rod of length L and mass m with its axis at its center is m*L^2/12.
That's true, but in this problem the rod has negligible mass so we ignore its moment of inertia.
Does m = 2M, the mass of just the spheres, or m = 5M, the mass of the whole rod-spheres-bug system?
I'm not sure what you're asking. For this problem, you need the moment of inertia of the system, which includes the two spheres (each with mass M) and the bug (mass 3M).
 
  • #13
That's true, but in this problem the rod has negligible mass so we ignore its moment of inertia.
But the only way to find the angular acceleration is to use the moment of inertia of the rod-spheres-bug system, which requires me to... Wait a minute...

The two spheres make the density of the system non-uniform...

Which means I need to calculate the second moment of inertia manually, right?

I'm not sure what you're asking. For this problem, you need the moment of inertia of the system, which includes the two spheres (each with mass M) and the bug (mass 3M).
You answered this one pretty well, thanks.
 

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