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Moment of Inertia and two uniform disks

  • #1
1. Homework Statement

Two uniform disks are connected by a light inextensible string supported by a massless pulley on a frictionless axis. (Very ideal LOL) The string is attached to a point on th circumference of disk A and wound around the disk B like a yo-yo.

Their moment of inertia does not equal 1/2mr^2 and the disks do not have uniform radial density.

For each disk A and B, the following are true:

I = .0254024 kgm^2
m = 1 kg
r = .2m
height = 1.6 m
g = 9.8 m/s^2

Which disk will reach the floor fit and what is the time interval for this disk to reach the floor?

2. Homework Equations

t = sqrt(2x/a) (Maybe?)


3. The Attempt at a Solution

As the spinning object would have a slower decent, object A would hit the ground first. (I think)

Past this, I'm totally lost. Anyone mind handing me a flashlight?
 

Answers and Replies

  • #2
Doc Al
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2. Homework Equations

t = sqrt(2x/a) (Maybe?)
But what's the acceleration?


3. The Attempt at a Solution

As the spinning object would have a slower decent, object A would hit the ground first. (I think)
What determines the acceleration of each disk? What forces act?
 
  • #3
But what's the acceleration?
That would be gravity, correct?


What determines the acceleration of each disk? What forces act?
Disk A (the one tethered to the string) would have g pulling down and B pulling it up.
Disk B (the yo-yo) would have g pulling down, A pulling it up, and centrifugal force (though it isn't really a force, so maybe I'm wrong)

Am I stumbling the right direction?
 
  • #4
Doc Al
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That would be gravity, correct?
Do you mean that a = g? No. (The acceleration would equal g if the objects were in free fall, but they're not.)
Disk A (the one tethered to the string) would have g pulling down and B pulling it up.
Disk B (the yo-yo) would have g pulling down, A pulling it up, and centrifugal force (though it isn't really a force, so maybe I'm wrong)
When analyzing the forces on A, forget about B. It's the string that pulls up on A.

I'll forget you ever said the phrase "centrifugal force". :yuck:

Again: What forces act on A? On B?
 
  • #5
Oh no.

*tries not to be obviously deficient*

Ok, A has g pulling down and the string pulling up.
B has g pulling down, the string pulling up, and um...something else having to do with the spin? o.o Torque maybe? (Please don't make fun of me :D)
 
  • #6
Doc Al
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Ok, A has g pulling down and the string pulling up.
Good. The weight acts down; the string tension acts up.
B has g pulling down, the string pulling up, and um...
The same two forces act on B.

Applying Newton's 2nd law will give you one equation.
something else having to do with the spin? o.o Torque maybe?
So far, we've just analyzed the translational motion. The same forces will create a torque on B. Analyze the torque on B and apply Newton's 2nd law for rotation.
 
  • #7
I have the same problem. I understand Tr=Ia/r (a/r=alpha) so you get mg-T= ma and T=Ia/r^2 so you get a=mg/(m+I/r^2), but when i put that a into sqrt(2h/a) i didn't get the right answer, perhaps i'm doing something wrong
 

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