Rotational dynamics Find acceleration

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The discussion focuses on solving a rotational dynamics problem involving two wheels and a hanging mass. The key equations derived include torque relationships for both the smaller and larger wheels, incorporating tensions that produce opposing moments. The user initially struggles with having more unknowns than equations but eventually realizes that by manipulating the equations, the tensions can be eliminated, allowing for the acceleration to be solved. The final insight emphasizes that changes in belt tension do not affect the overall dynamics of the system. The problem is resolved successfully with the correct application of rotational dynamics principles.
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Homework Statement


http://imgur.com/eYoa8Ut

Homework Equations


Torque=rt=I(alpha)
t-mg=-ma
alpha = a/r

The Attempt at a Solution


t-mg=-ma is the equation for the hanging mass.
To find how much this turns the small wheel,
rt=I(alpha) I and r are known
rt=Ia/r

I'm not sure what to do from here. If I try and solve for a and t now, it gives an incorrect answer. I think this is due to me not doing anything with the larger wheel yet, but I'm not sure what to do with it. I know that the tension in the upper and lower part of the belt are not equal, and both produce torque of their own. Tupper should produce a clockwise moment, Tlower should produce a counterclockwise moment.
 
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ForceFail said:
Tupper should produce a clockwise moment, Tlower should produce a counterclockwise moment.
Quite so. So include these in your torque equation for the smaller wheel and write down the torque equation for the larger wheel.
 
So for the smaller wheel:
tr+(Tl)R-(Tu)R=I(alpha)
Larger wheel gives:
(Tu)(Rlw)-(Tl)(Rlw)=I(alpha)

So now I have 2 equations and four unknowns. t,(Tl),(Tu),(alpha)
I can write t=mg-ma and alpha as a/r, Does alpha = a/r for the small wheel use the smaller or larger radius? Since its a tangential I would assume the larger.
That gives
(mg-ma)r+(Tl)R-(Tu)R=Ia/R
and
(Tu)(Rlw)-(Tl)(Rlw)=Ia/(Rlw)

If this is right I'm now down to 3 unknowns and only 2 equations.
 
ForceFail said:
So for the smaller wheel:
tr+(Tl)R-(Tu)R=I(alpha)
Larger wheel gives:
(Tu)(Rlw)-(Tl)(Rlw)=I(alpha)

So now I have 2 equations and four unknowns. t,(Tl),(Tu),(alpha)
I can write t=mg-ma and alpha as a/r, Does alpha = a/r for the small wheel use the smaller or larger radius? Since its a tangential I would assume the larger.
That gives
(mg-ma)r+(Tl)R-(Tu)R=Ia/R
and
(Tu)(Rlw)-(Tl)(Rlw)=Ia/(Rlw)

If this is right I'm now down to 3 unknowns and only 2 equations.
Look carefully at how Tu and Tl appear in these equations. Notice anything?
 
haruspex said:
Look carefully at how Tu and Tl appear in these equations. Notice anything?

I see, multiplied equation 1 by R(lw) and equation 2 by R and added them. Then Tu and Tl cancel and you can solve for a. Got the right answer Thanks.
 
ForceFail said:
I see, multiplied equation 1 by R(lw) and equation 2 by R and added them. Then Tu and Tl cancel and you can solve for a. Got the right answer Thanks.
Good.
Note what this means physically. Suppose the belt is elastic. Making it tighter will increase both tensions equally, but won't affect the answer to the question.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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