Atwood's with a cylinder - Rotational Motion

[Null_Void]

Homework Statement

Pic below containing question

Relevant Equations

$ωR = 2v$
$I_{cylinder} = mR^2/2$

$mg- T = ma_1$ (for the block)
$mg - T = ma_2$ (for the cylinder)

I am confused about three things:

(i) To prove that accelerations are same, in the book it is given:
$mg−T=ma_1$ (for the block)
$mg−T=ma_2$ (for the cylinder)
$(a_1 = a_2 = a)$
And thus the accelerations should be same. I can see that the same forces are acting on the bodies and that the cylinder will unwind but I'm confused since the masses are same and I'm not sure about the directions. How can I conclude what would happen to the system when released from rest?

(ii) For the non-slip condition, I know that the string and the point on the rim of the cylinder must have the same velocity.
When I tried to calculate it:
$v−ωR=−v$
$ωR=2v$
I assumed that the string must move upwards with velocity v due to the block. The torque is clockwise, hence we get the term on the left. Though this gives the right answer, I'm not sure if my method is right?

(iii)When I try to calculate accelerations by using forces, from (ii) we get

$αR=2a$
From our original force equation,

$mg−T=ma$

Writing torque equation about the COM of cylinder,
$TR=(mR^2+mR^2/2)α$

Combining all equations,
$Τ=3ma$
$a=g/4$

But the answer is $a=g/2$
Where am I going wrong?

Any help is appreciated!
Thank you

 

[kuruman]

What block? What cylinder? Please provide the description of the physical situation, not just the solution given to you.

 

[Null_Void]

What block? What cylinder? If you have a massive cylinder why is its moment of inertia ignored? Please provide the description of the physical situation, not just the solution given to you.

You have a normal block of mass $m$ hanging at one end of the string and a cylinder of radius $R$ and moment of inertia $mR^2/2$ about it's center of mass on which the string is wound. The pulley and string are massless and ideal.
See pic for reference.

 

[haruspex]

And thus the accelerations should be same. I can see that the same forces are acting on the bodies and that the cylinder will unwind but I'm confused since the masses are same and I'm not sure about the directions. How can I conclude what would happen to the system when released from rest?

You have to trust F=ma. It does not care about the rotation.

(ii) For the non-slip condition, I know that the string and the point on the rim of the cylinder must have the same velocity.
When I tried to calculate it:
$v−ωR=−v$
$ωR=2v$
I assumed that the string must move upwards with velocity v due to the block. The torque is clockwise, hence we get the term on the left. Though this gives the right answer, I'm not sure if my method is right?

Yes, that works.

(iii)When I try to calculate accelerations by using forces, from (ii) we get

Writing torque equation about the COM of cylinder,
$TR=(mR^2+mR^2/2)α$

You have used the moment of inertia about a point on the rim, but the torque TR does not act about that point.

 

[kuruman]

I see now. I missed the second picture.

To rephrase what @haruspex already said, the net torque on the cylinder pulley is zero because the tension is the same on both sides, therefore the cylinder pulley has zero angular acceleration.

 

[Null_Void]

You have to trust F=ma. It does not care about the rotation.

To expand on my confusion, if the string were connected in such a way as the block (on the same line as the COM), we know that there would be no acceleration. But how do we now know that both will accelerate downwards specifically when we encounter the problem for the first time.
Why can't I assume the equations to be :

$mg - T = ma_1$
$T - mg = ma_2$

Yes, that works.

Just for confirmation, why doesn't the cylinder's velocity affect the net velocity of the point on the string. Why is the point's velocity influenced only by the block?

You have used the moment of inertia about a point on the rim, but the torque TR does not act about that point.

I'm extremely sorry, I see my mistake now.
$TR = (mR^2/2)α$
$T = ma$
Which gives the right answer $a = g/2$. Thanks a lot for the help.

 

[haruspex]

To expand on my confusion, if the string were connected in such a way as the block (on the same line as the COM), we know that there would be no acceleration. But how do we now know that both will accelerate downwards specifically when we encounter the problem for the first time.
Why can't I assume the equations to be :

$mg - T = ma_1$
$T - mg = ma_2$

You can, but in the first of those you are taking down as positive and in the second you are taking up as positive. So when we add them and get $a_1=-a_2$ we still discover that the accelerations are the same in magnitude and direction.

Just for confirmation, why doesn't the cylinder's velocity affect the net velocity of the point on the string. Why is the point's velocity influenced only by the block?

It is related to both, as you used in your equation in part (ii). It is unhelpful to think of it in terms of causality. The different motions merely have to satisfy some consistency relations.

 

[haruspex]

I see now. I missed the second picture.

To rephrase what @haruspex already said, the net torque on the cylinder is zero because the tension is the same on both sides, therefore the cylinder has zero angular acceleration.

Are you confusing the massive cylinder with the massless pulley?

 

[kuruman]

Are you confusing the massive cylinder with the massless pulley?

Yes. The cylinder has only one tension acting on it.

 

[Null_Void]

You can, but in the first of those you are taking down as positive and in the second you are taking up as positive. So when we add them and get $a_1=-a_2$ we still discover that the accelerations are the same in magnitude and direction.

Thank you very much. I get it now.

It is related to both, as you used in your equation in part (ii). It is unhelpful to think of it in terms of causality. The different motions merely have to satisfy some consistency relations.

If you don't mind me asking, could you explain this a bit more in detail. My main confusion with regard to this is that,
Say you have a cylinder on a floor and it is given some velocity $v$. Since I know that the ground can't move, for the non-slip condition, I know that the net velocity at the bottom should be zero. In our case since we do not have a fixed surface such as the ground, how can we be sure of equating the velocities like we did and choosing the actual velocity of the point?

 

[haruspex]

In our case since we do not have a fixed surface such as the ground

Whether one surface is fixed is irrelevant. What matters is that the two surfaces in contact have the same velocity component in the contact plane.

 

[Null_Void]

Whether one surface is fixed is irrelevant. What matters is that the two surfaces in contact have the same velocity component in the contact plane.

But how can we be sure that the net velocity will be $-v$? I initially got the answer by simply assuming. That's why this is bugging me.
Thank you

 

[haruspex]

But how can we be sure that the net velocity will be $-v$? I initially got the answer by simply assuming. That's why this is bugging me.
Thank you

We know that the two masses have the same acceleration at all times and the system is released from rest. They start with the same velocity, so continue to have the same velocity.
The upward velocity of the ascending string section must be -v.