Torque and force problem -- Atwood's Machine Motion

[Fascheue]

Homework Statement

Solve problem 8.1, but using force and torque rather than conservation of energy. (Image of 8.1 attached)

Homework Equations

T = f x d

f = ma

The Attempt at a Solution

I think I might have to find the torque on the disk, which I’m guessing would be 2mgr - mgr, but I’m not quite where to go from here or if that is even correct.

 

[RedDelicious]

How is the net torque on the disk related to its motion? What was your motivation for finding the torque to begin with?

 

[Fascheue]

How is the net torque on the disk related to its motion? What was your motivation for finding the torque to begin with?

I know torque has some relation to acceleration. I think it might be T = Iw’, and since the pulley mass is a disk I = 1/4mr^2. If I plug in the torque gotten from the previous equation, mgr, that gives me 4g/r for w’. I think I should be able to get the linear acceleration of the disk from this, which should be equal to the acceleration of the system, but I’m not quite sure what that formula is.

 

[haruspex]

I = 1/4mr^2

Not 1/4. Try again.

should be able to get the linear acceleration of the disk from this

Yes. If a disc radius r rotates through an angle θ, through what distance does a point on the rim move?

 

[Fascheue]

Not 1/4. Try again.

Yes. If a disc radius r rotates through an angle θ, through what distance does a point on the rim move?

Oops, I must’ve found the moment of inertia for the wrong axis. I believe it should be 1/2mr^2Would the point move a distance rθ?

 

[haruspex]

Would the point move a distance rθ?

Right, and for circular motion you can find the relationship between linear velocity and angular velocity by differentiating, and again for accelerations.

 

[Fascheue]

Right, and for circular motion you can find the relationship between linear velocity and angular velocity by differentiating, and again for accelerations.

I think that should make the relationship a = rθ’’

Using the equation T = Iθ’’ where I is 1/2mr^2 and T is mgr, I got 2g/r for θ’’.

Then converting to linear acceleration using the formula a = rθ’’ gave me a = 2g.

 

[BvU]

That would be twice as much as free fall; doesn't that look a bit weird ?

A tip: what do the FBDs for $m$ and the $2m$ look like ? Then make the link with the $\tau = I\alpha$ for the disk.

 

[Fascheue]

That would be twice as much as free fall; doesn't that look a bit weird ?

A tip: what do the FBDs for $m$ and the $2m$ look like ? Then make the link with the $\tau = I\alpha$ for the disk.

I now see why 2g doesn’t make sense, but I can’t find my mistake.

The torque seems like it should be 2mgr - mgr = mgr

I = 1/2mr^2

T = Iθ’’

mgr = 1/2mr^2θ’’

θ’’ = 2g/r

a = rθ’’

a = 2g

 

[BvU]

Let me then convert my tip into a direct question:

what do the FBDs for $m$ and the $2m$ look like

 

[Fascheue]

Let me then convert my tip into a direct question:

For 2m: 2mg -T

For m: mg - T

Where + is down

 

[BvU]

If the two T are equal, there is no net torque on the disk

 

[Fascheue]

If the two T are equal, there is no net torque on the disk

So would they be

m1: mg - T1

m2: 2mg - T2

And then the forces on the disk be the tensions in the rope?

Also, I thought tension was supposed to be the same everywhere unless the rope had mass, is that then incorrect?

 

[BvU]

Yes, that is indeed not correct where the rope is in contact along the perimeter of the disk. Think friction.

 

[BvU]

And then the forces on the disk be the tensions in the rope?

Yes. The third and fourth force on the disk are gravity and the suspension.

Now count your unknowns and your equations so see if you can solve...

 

[haruspex]

I thought tension was supposed to be the same everywhere unless the rope had mass, is that then incorrect?

That's because you are accustomed to questions which specify massless pulleys (or static force balances).

 

[Fascheue]

Yes. The third and fourth force on the disk are gravity and the suspension.

Now count your unknowns and your equations so see if you can solve...

Hmm, I keep getting 2/3g but it looks like my book is saying that the answer is 2/7g

I had t = T2r - T1rwhere t is torque and the T’s are tensions. I don’t believe the other forced add torque because they are acting on the center of the disk.

Then I = 1/2mr^2

t = Iθ’’

T2r - T1r = 1/2mr^2θ’’

I then found that T1 = 1/2T2 by setting the force acting on the second mass by the second mass equal to the force acting on the first mass divided by the first mass.

I now have 1/2T2r = 1/2mr^2θ’’

T2 = mrθ’’

θ’’ = T2/mr

a = rθ’’

a = T2/m

Which should be equal to the forces on the second mass divided by the second mass

(2mg - T2)/2m= T2/m

(2mg - T2)/2 = T2

mg - 1/2T2 = T2

2/3mg = T2

a = 2/3g

 

[haruspex]

found that T1 = 1/2T2 by setting the force acting on the second mass by the second mass equal to the force acting on the first mass divided by the first mass.

But those tensions are not the only forces acting on those masses. The net forces must be in proportion to the masses.

 

[Fascheue]

But those tensions are not the only forces acting on those masses. The net forces must be in proportion to the masses.

I didn’t use just the tensions for those forces

I had (2mg - T2)/2m = (mg-T1)/m

Multiplied by m

(2mg - T2)/2 = mg - T1

mg - 1/2T2 = mg - T1

mg cancels out

-1/2T2 = -T1

T1 = 1/2T2

 

[haruspex]

I had (2mg - T2)/2m = (mg-T1)/m

What is the left hand side of that in terms of a? Likewise, the right hand side? Be careful.

 

[Fascheue]

What is the left hand side of that in terms of a? Likewise, the right hand side? Be careful.

I added the negative to one side of the equation, finished solving, and ended up with the correct answer.

Thanks everyone for the help.