Question regarding block and pulley system with inertia

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Homework Help Overview

The problem involves a block and pulley system where two blocks are connected by a string over a disk pulley. The original poster questions the equation for acceleration provided in a textbook, suggesting a discrepancy in the terms involving the moment of inertia.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the acceleration of the blocks and the tensions in the system, while questioning the validity of a term in the textbook's equation. Some participants check the dimensional consistency of the terms involved, while others confirm the original poster's approach to the problem.

Discussion Status

Participants are actively engaging with the original poster's reasoning, providing feedback on the dimensional analysis and confirming the correctness of the approach taken. There is a shared exploration of the implications of the moment of inertia in the context of the problem.

Contextual Notes

There is an indication that the problem is sourced from a textbook used in an engineering course, which may influence the expectations regarding the accuracy of the equations presented.

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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