Blocks falling while attatched to pulley

[Jamie_Pi]

Homework Statement

The two blocks, m1 = 3.3 kg and m2 = 4.8, in the figure below are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.64 N · m. If the blocks are released from rest, how long does it take the 4.8 kg block to reach the floor from a height of h = 1.0 m? (Note: If your random numbers do not create movement between the masses enter 0 for your answer.)

Homework Equations

Torque=Force*Radius
Torque=moment of inertia*angular acceleration
angular acceleration = linear acceleration/radius
s=1/2*linear acceleration*time^2

The Attempt at a Solution

I tried to calculate the torque on the pulley and convert that to linear acceleration, and find the time from that.
Torque total= m*g*r-friction
Torque total=4.8*9.8*0.06-4.8*9.8*0.06-0.64
Torque total=0.4004
moment of inertia=1/2*mass*radius^2
moment of inertia=0.0036
0.4004=0.0036*angular acceleration
angular acceleration=111.22
111.22=linear acceleration/0.06
linear acceleration=6.6733
s=3.337*t^2
1=3.337*t^2
t=sqrt(1/3.337)
t=0.5441

I did something wrong, and I'm not sure what. I'm not very good with rotation and torque, though, so I don't know where my mistake could be. The real answer is 2.12 seconds.

 

[haruspex]

Torque total=4.8*9.8*0.06-4.8*9.8*0.06-0.64

I assume you meant 4.8*9.8*0.06-3.3*9.8*0.06-0.64

Torque total=0.4004

Seems a bit too much. Please post your working.

0.4004=0.0036*angular acceleration

You cannot take the entire torque and apply it all to the pulley. Some of it will go into accelerating the masses.
Either:
- Assign distinct variables to the two tensions and write the three acceleration equations for the two masses and the pulley,
or:
- represent the masses in the torque equation (as mr²)

 

[Jamie_Pi]

I assume you meant 4.8*9.8*0.06-3.3*9.8*0.06-0.64

Seems a bit too much. Please post your working.

You cannot take the entire torque and apply it all to the pulley. Some of it will go into accelerating the masses.
Either:
- Assign distinct variables to the two tensions and write the three acceleration equations for the two masses and the pulley,
or:
- represent the masses in the torque equation (as mr²)

Ok, I re did it and I must have done some math wrong before.
Torque=0.242

If I understand your suggestions correctly, I think representing the masses as part of the pulley is the best idea, so I'll do:
moment of inertia= 1/2(2+4.8+3.3)*0.06^2

So from there,
torque=moment of inertia * angular acceleration
0.242=1/2*(2+4.8+3.3)*0.06^2*a/0.06 ('a' being linear acceleration)
a=(2*0.242)/(2+4.8+3.3+0.06)
a=0.476
s=1/2a*t^2 ('s' is 1)
t=sqrt(2*1/a)

this gives me t=2.049*, which is a bit off, but better. I must have misunderstood what you meant about representing the mass in the torque equation, could you explain that a little bit more? I feel like I'm really making progress on this one.

*this is an edit, earlier I messed up my math and had the incorrect number here.

 

[haruspex]

representing the masses as part of the pulley is the best idea, so I'll do:
moment of inertia= 1/2(2+4.8+3.3)*0.06^2

The masses are not disks rotating about their centres. All of the mass acts at the same radius as far as the pulley is concerned.

 

[Jamie_Pi]

The masses are not disks rotating about their centres. All of the mass acts at the same radius as far as the pulley is concerned.

Ok, that makes sense. I'm not sure what to represent them as, though, because they aren't actually spinning at all. I tried representing them as points rotating around the central axis, and that got me somewhat close:

torque=(1/2*2*0.06^2+(4.8+3.3)*0.06^2)*a/0.06
which I simplified into
torque=(4.8+3.3)*0.06*a

and plugging this back into the same thing:
0.242=0.06*8.1*a
a=0.4979
time=sqrt(2/0.4979)

time=2.004 seconds, which is pretty close.

 

[haruspex]

torque=(4.8+3.3)*0.06*a

You left out the pulley.

 

[Jamie_Pi]

You left out the pulley.

Well, since the pulley is 2kg and it is multiplied by 1/2, it became 1 and wasn't really doing anything.

Ahh, but of course! It's being added not multiplied.
torque=0.06*(8.1+1)*a
0.242/(0.06*9.1)=a
a=0.4432 (looks good so far)
time=sqrt(2/0.4432)

time= 2.12
Yay! Thanks a bunch! Again.