Question regarding block and pulley system with inertia

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The discussion centers on a problem from Mosca and Tipler regarding a block and pulley system, where participants analyze the equations for acceleration and tension. The original poster believes the book's equation for acceleration is incorrect, specifically questioning the term (I/r)^2 used in the book. They derive the correct acceleration as a=(m2/(m2+m1+I/r^2))*g, confirming that the term I/r^2 is appropriate for dimensional consistency. Other contributors agree with this assessment and clarify the relationships between tension, torque, and acceleration in the system. The consensus is that the book's notation is misleading, prompting a suggestion to provide feedback to the authors.
jcruise322
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This problem is an example from Mosca and Tipler, 9-13, 6th edition. I believe the books equation for the acceleration of the system is incorrect according to my work... Anyway, here goes

1. Homework Statement

"Two blocks are connected by a string that passes over a disk pulley of radius R and moment of inertia I. The block of mass m1 slides on a frictionless, horizontal surface; the block of mass M2 is suspended from a string over the pulley. Find the acceleration of the blocks and the tensions. The string does not slip on the pulley."

Homework Equations


Non slip, so: a=r*α
For pulley: r(T2-T2)=I*α
For block 2: m2g-T2=m2*a m2=mass two
For block 1: T1=m1*a m1=mass one

The Attempt at a Solution


[/B]
Well, for the pulley: T2-T1=(I*α)r, and therefore: T2-T1=(I*a)/r^2 (substituting a/r=α).

I added the above equation to the two others so the tensions would cancel:

so: m2*g=(m2+m1+I/r^2)*a

so: a=(m2/(m2+m1+I/r^2))*g
The books says the the third term in the middle parentheses is (I/r)^2...how is this possible? Shouldn't it be I/r^2? Any input would be greatly appreciated! Thanks! :)
 
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Check the dimensions - you can only add a mass to m1 and m2, so a term ##I / r^2## is acceptable, a term ##(I/r)^2## is surely not.
 
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jcruise322 said:
Non slip, so: a=r*α
For pulley: r(T2-T2)=I*α
For block 2: m2g-T2=m2*a m2=mass two
For block 1: T1=m1*a m1=mass one

Everything looks correct.

jcruise322 said:
Well, for the pulley: T2-T1=(I*α)r, and therefore: T2-T1=(I*a)/r^2 (substituting a/r=α).

I added the above equation to the two others so the tensions would cancel:

so: m2*g=(m2+m1+I/r^2)*a

so: a=(m2/(m2+m1+I/r^2))*g

You're right, from the torques acting on the pulley you get:

T2-T1=[(Mass of the pulley)*R*α]/2 = (I*α)/R

and from the no slip condition you know that α=a/R therefore

T2-T1=[(Mass of the pulley)*a]/2 = (I*a) / R^2

and solving for a, you get:
a=(m2/(m2+m1+I/R^2))*g, which is what you did.
 
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BvU said:
Check the dimensions - you can only add a mass to m1 and m2, so a term I/r2I / r^2 is acceptable, a term (I/r)2(I/r)^2 is surely not.
BvU is right!
 
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Awesome, thanks guys! I will send Mosca and Tipler a "tip" on how they can rewrite their physics book haha. We use it for engineering here at UW Seattle :) Again, really appreciated :) :)
 
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