2 masses suspended over a pulley, finding a mistake

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The discussion revolves around a physics problem involving two masses suspended over a pulley, where the user identifies a mistake in their calculations regarding the moment of inertia. The user initially derives the acceleration 'a' using Newton's second law and the relationship between tension and moment of inertia. They realize that their solution incorrectly uses a single moment of inertia 'I' instead of the correct value of '2I'. Other participants confirm the user's calculations are correct but point out the professor's error in the problem statement. The conversation highlights the importance of accuracy in physics problems and acknowledges that mistakes can occur from both students and educators.
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Homework Statement


http://i.imgur.com/1j19V0n.jpg
Two masses are suspended from a pulley as on the pic. I have pulleys radius R and moment of inertia I and masses m1 m2.


Homework Equations





The Attempt at a Solution



So 2nd law of motion:

N_{1} - m_{1}g=m_{1}a
m_{2}g-N_{2} = m_{2}a
(N_{2}-N_{1})R=Iε=\frac{Ia}{R}

I get N's from 3rd equation and add two first ones

N_{2}-N_{1}=\frac{Ia}{R^{2}}
g(m_{2}-m_{1}) + N_{1} - N_{2} = (m_{1}+m_{2})a

sub N's to second equation

g(m_{2}-m_{1}) - \frac{Ia}{R^{2}} = (m_{1}+m_{2})a
g(m_{2}-m_{1}) = (m_{1}+m_{2})a + \frac{Ia}{R^{2}}
g(m_{2}-m_{1}) = a(m_{1}+m_{2}+\frac{I}{R^{2}})
\frac{g(m_{2}-m_{1})}{m_{1}+m_{2}+\frac{I}{R^{2}}} = a

Now, it's wrong as there should be 2I and not I. Where did I make a mistake?
 
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aquance said:
\frac{g(m_{2}-m_{1})}{m_{1}+m_{2}+\frac{I}{R^{2}}} = a

Now, it's wrong as there should be 2I and not I. Where did I make a mistake?

Your solution is correct.

ehild
 
Uh, thanks, it seems that prof made a mistake.
 
aquance said:
Uh, thanks, it seems that prof made a mistake.

Everybody can make mistakes... :biggrin:

ehild
 
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