Homework Help: 2 hanging masses on Pully with friction time to reach floor?

1. Apr 23, 2010

charan1

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

http://session.masteringphysics.com/problemAsset/1070551/3/12.P70.jpg

The two blocks in the figure are connected by a massless rope that passes over a pulley. The pulley is 15 cm in diameter and has a mass of 2.6 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.46 N-m.

If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

2. Relevant equations

Tnet=I$$\alpha$$

T=FD

F=ma

$$\alpha$$=a/r

Moment of inertia of a disc:
I=.5mr2

g is gravity!
3. The attempt at a solution

T1 is tension of the left mass, T2 is tension of the right mass
m1 is mass of left mass, m2 is mass of right mass
T1=-m1a+m1g
T2=m2a+m2g

$$\Sigma$$T= TT1 - TPully friction - TT2

I$$\alpha$$=TT1 - TPully friction - TT2

Ia/r=(-m1a+m1g)r - .46 - (m2a+m2g)r

Ia/r= -m1ar + m1gr - .46 -m2ar -m2gr

Ia = -m1ar2 + m1gr2 - .46r -m2ar2 -m2gr2

Ia + m1ar2 + m2ar2 = m1gr2 - .46r -m2gr2

a(I + m1r2 + m2r2) = m1gr2 - .46r -m2gr2

a = (m1gr2 - .46r -m2gr2) / (I + m1r2 + m2r2)

I=.5mr2

I = (.5)(2.6)(.075)2
I = .0073125

m1 = 4kg
m2 = 2kg
r = .075m
g=9.8

a = ((4)(9.8)(.0752) - .46(.075) -(2)(9.8)(.075)2) / (.0073125) + (4)(.075)2 + (2)(.075)2

a = (.2205 - .0345 - .11025) / (.0073125 + .0225 + .01125)

a = .07575/.0410625

a = 1.8447

yf=yi+vit+.5at^2

0=1 + 0 + .92235t^2

t=1.04124 seconds

is this correct?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 23, 2010

jdc15

Just double checked with a slightly different method. Looks right to me.

3. Apr 23, 2010

charan1

Second opinion please anyone!

Just want to make sure! Dont have many tries left on the problem

4. Apr 23, 2010

jdc15

If you wanted to be pedantic, to be more correct you would round to two significant figures (1.0 rather than 1.04124). Also, at this step, probably a typo but there should be a negative sign or have the 1 moved to the other side:

0=1 + 0 + .92235t^2

change to

0=-1 + 0 + .92235t^2

The answer should be right though.