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Moment of Inertia (Atwood's Machine)

  • Thread starter garyljc
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Question : The masses of blocks A and B are given M_a and M_b , respectively, the moment of inertia of the wheel about its axis is I , and the radius of the semicircle in which the string moves is R. Assume that there is no slippage between the wheel and the string.
Also M_a > M_b

a.) Find an expression for the magnitude of the linear acceleration of block B in terms of the linear acceleration a_a ( acceleration of block A ) .

My attempt :
Using Newton's law of motion
we have
g.M_a - T1 = M_a . a_a
T2 - g.M_b = M_b. a_b , where a_b is the acceleration of block B
and torque of the pulley : T1.R - T2.R = I (alpha) , where alpha is the angular acceleration

Am I heading the right direction? If so, how could i eliminate both tensions T1 and T2 to get what the question wants ?
 

Answers and Replies

  • #2
Am I heading the right direction? If so, how could i eliminate both tensions T1 and T2 to get what the question wants ?
You're almost there but like all classical Newton's problems you need some constraints to get you all the way there.

Currently you have too many unknown variables, but a couple statements can help you out.

1) The system as a whole (that is block A and block B) move together (assuming a taught string) and so aa = ab = a

2) You will need a constraint that eliminates [tex]\alpha[/tex] with a classical relation to acceleration and radius. I believe, (but am not positive) you can use [tex]\alpha[/tex]= a/r

Now as for getting rid of your tensions, if you re-arrange your torque equation to look like this: [tex]\tau[/tex] net = (T1 - T2)r = I[tex]\alpha[/tex], then you can see that a simple subtraction of your first two equations can be used to substitute the tensions away.

-eq1 - eq2 => (T1 - T2) + (Mbg - Mag) = (-Ma - Mb)a

you can then solve for your (T1 - T2) term and plug it into equation 3. All should work out. Hope this helps,

cheers
 

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