# Tension and Torque homework challenge problem.

## Homework Statement

A 4KG mass is connected by a weightless cord to a 3 kg mass on a smooth surface. The pully rotates about a frictionless axle and has a moment of inertia of 0.5 kgm^2 and radius of 0.3m. Assuming that the cord does not slip on the pully, find a.) the acceleration of the two masses and b.) the tensions T1 and T2.

Click the image below to expand it to see what this looks like.

http://img98.imageshack.us/img98/1032/47770145rn2.th.png [Broken]

## Homework Equations

T1=39.24-ma
T2=3a
Torque1=0.3T1
Torque2=0.3T2
I=mr^2
Net Torque=Ialpha

## The Attempt at a Solution

I started by finding the mass(M) of the pully.
0.5=0.09M
M=5.56
I then used the net torque equation, substituting a/r for alpha.
nettorque=1.67a
I then summed my torques to this.
torque1-torque2=1.67a
0.3T1-0.3T2=1.67a
0.3(T1-T2)=1.67a
T1-T2=5.56a
Next I substituted my tensions in.
(39.24-4a)-3a=5.56a
39.24=12.56a
3.12a

I figured that now I could substitute the acceleration back in my original tension equations to find the tensions.
T2=3(3.12)
T2=9.37
T1=39.24-12.48
T1=26.76

There is no answer in the back of the book for this one, and I want somebody to check(rather than skim) my work and tell me if I have come to the right conclusion. If not, please tell me what I have done wrong. It's important that I can do this kind of equation for this week's test.

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Looks good to me except for this first step:
I started by finding the mass(M) of the pully.
0.5=0.09M
M=5.56
(1) You don't need the mass of the pulley, only the rotational inertia, which is given.
(2) Your equation for rotational inertia (I = mR^2) is incorrect; if you model the pulley as a uniform disk, I = 1/2mR^2.

Luckily, you made no further use of this calculation, so the rest of your work is fine.