Uniform disk tension problem

  • #1
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I have a diagram for this at http://snipurl.com/diagr [Broken]

There is a uniform disk with radius R, mass M, it has a string wrapped around it and is attached to a fixed rod.
Part 1: Show the tension in the string is 1/3 the weight of the disk

what i have done for this is

[tex] Torque=Fd [/tex]
and setting the force equal to Tension
[tex] Torque=TR [/tex]
then [tex] \sum Torque=I\frac{a}{R}[/tex] and i used [tex]I=\frac{1}{2}MR^2[/tex]for the moment of inertia

Then [tex] TR=\frac{1}{2}MRa[/tex]
[tex]T=\frac{1}{2}Ma[/tex]
and then acceleration is just g, so i get the tension equals 1/2 the weight. where have i gone wrong????
 
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Answers and Replies

  • #2
ehild
Homework Helper
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thenewbosco said:
I have a diagram for this at http://snipurl.com/diagr [Broken]

There is a uniform disk with radius R, mass M, it has a string wrapped around it and is attached to a fixed rod.
Part 1: Show the tension in the string is 1/3 the weight of the disk

what i have done for this is

....
[tex]T=\frac{1}{2}Ma[/tex]
and then acceleration is just g, so i get the tension equals 1/2 the weight. where have i gone wrong????

The acceleration of the centre of mass is not just g, but it is determined by the resultant of all forces acting on the pulley.

ehild
 
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  • #3
Andrew Mason
Science Advisor
Homework Helper
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thenewbosco said:
where have i gone wrong????
Your method is correct except for equating a to g. The forces are Mg and T, so Mg-T=ma. You have found that T = Ma/2 so T = Mg/3. Here is another way of looking at it:

The weight is Mg. The tension is T. The net force provides the acceleration of the wheel:

(1) Mg - T = Ma

To find the acceleration of the center of mass:
The torque of the center of mass about the point of rotation (point of contact between wheel and string) is MgR. That torque divided by the moment of inertia of the wheel about that contact point provides the angular accleration of the centre of mass = a/R.

So:

(2)[tex]MgR/I = a/R [/tex]

The moment of inertia about that point is NOT the same as the moment of inertia about the center. Use the parallel axis theorem: [itex]I = I_{centre} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2[/itex]

Plugging that into (2) results in a = 2g/3

Then it is just a matter of plugging that into (1) to get:

T = Mg(1 - 2/3) = Mg/3

AM
 
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