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

[charan1]

Homework Statement

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?

Homework Equations

T_net=I$$\alpha$$

T=FD

F=ma

$$\alpha$$=a/r

Moment of inertia of a disc:
I=.5mr²

g is gravity!

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= T_T1 - T_{Pully friction} - T_T2

I$$\alpha$$=T_T1 - T_{Pully friction} - T_T2

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?

 

[jdc15]

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

 

[charan1]

Second opinion please anyone!

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

 

[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.