- #1
jcruise322
- 36
- 1
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."
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
[/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! :)
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! :)