Torque/acceleration with masses on pulley

1. Oct 28, 2004

vtech

Wer are given two objects of mass m1 = .20 kg and m2 = .67 kg, which are connected by a massless cord that is wrapped around a uniform disk of mass M = 0.50 kg and radius R = 0.12 m. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system is released from rest.

(a) Find the magnitude of the acceleration of the blocks____m/s2
(b) Find the tension T1 in the cord at the left____ N
(c) Find the tension T2 in the cord at the right____N
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well so far i have three equations - clockwise motion
1) T2*R - T1*R = I*alpha = net torque where alpha = a/R for pulley
2) T2 - m2*g = m2*a for mass on pulley
3) -T1 + m1*g = m1*a for mass on pulley

Then I solve 2) and 3) for respective tensions T2, T1 and substitute into 1) and solve for "a" I mean, am I close here or no? Cuz I don't wanna write all of my solving out if I'm at the wrong starting point anyway...kinda messy.... but I'm getting an answer of

a = - 7.429m/s2 but I get it as wrong. Don't have actual answer. Negative? So I'm guessing goes in opposite direction...???

Without correct a can't solve for T1 & T2 ... for answer a) since magnitude is being asked, even if acceleration is negative answer is +ve value.

I'd really appreciate some input on this questions! Thanks in advance!

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2. Oct 28, 2004

vtech

can anyone show me how to include a figure/graph in my question area instead of putting in as an attachment??

3. Oct 29, 2004

vtech

can anyone help on this one?

surely there is someone that can help out with this problem? i understand the concept of the torque acting with the forces, but i am unsure as to how i can relate this with the tensions! help anyone??? i would really appreciate it!!!

4. Oct 29, 2004

Staff: Mentor

sign problems

You are mixing up the signs for the acceleration. You need to use a consistent sign convention. If you take up as positive, then the acceleration of m1 is up (+a), but the acceleration of m2 is down (-a). Rewrite equations 2 and 3.