# Question regarding block and pulley system with inertia

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1. Nov 30, 2015

### jcruise322

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. The problem statement, all variables and given/known data

"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."

2. Relevant 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

3. The attempt at a solution

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! :)

2. Nov 30, 2015

### BvU

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.

3. Nov 30, 2015

### (Ron)^2=-1

Everything looks correct.

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.

4. Nov 30, 2015

### (Ron)^2=-1

BvU is right!

5. Nov 30, 2015

### jcruise322

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 :) :)