# Pulleys, torque and tension

A 1.2 kg mass and a 7.1 kg mass are connected by a massless string over a pulley that is in the shape of a bicycle wheel (all mass is concentrated at the rim) having a radius 0.25 m and mass 4.6 kg. Determine the time for the masses to move 2.6 m starting from rest.

A.
2.714 sec
B.
0.223 sec
C.
1.55 sec
D.
1.68 sec
E.
1.077 sec

T1 = 7.1kg, T2 = 1.2kg

T1 - m1g = -m1a
T2 - m2g = m2a
T1R - T2R = I(alpha)

a=R(alpha)

I was just wondering if I set these equations properly. I substituted and got an answer similar to E except its off a little even though I didn't round. Thanks in adv!

## Answers and Replies

rl.bhat
Homework Helper
Hi PhrozenN, welcome to PF.
Unless you show your calculations we are not in a position to point out your mistakes.
So please show your calculations. By the way, which expression you have used for the moment of inertia for the pulley?

I = mR^2 since its a rim-like thingy

T1R - T2R = I(a/R)
m1(g-a)R - m2(g+a)R = mRa
a = 4.44 m/s^2

x = vt + 0.5at^2
t = 1.082

rl.bhat
Homework Helper
Check the calculation of a. I am getting the correct answer.

Oh, I did the calculations wrong
I got a = 4.48 so now it's 1.077

The equations are set properly right? Thanks btw