Net Torque Within a Mass and Pulley System

[MichaelKD]

Homework Statement

A counterweight of mass m = 4.00 kg is attached to a light cord that is wound around a pulley. The pulley is a thin hoop of radius R = 8.00 cm and mass M = 2.00 kg. The spokes have negligible mass. (a) What is the magnitude of the net torque on the system about the axle of the pulley?

Relevant Equations

tau=I*alpha=Fr, F=ma

I started by summing the forces and torques to get:
- ma = mg-T
- I*alpha=Tr

I then used a=alpha*r and I=Mr^2 to combine the equations and solved for angular acceleration equals 81.75rad/s^2. Plugging this back into a torque equation I got that the net torque is 1.04Nm. However, the problem says the answer is 3.14Nm (which is my answer times 3. Does anyone know where I went wrong?

 

[gneill]

Your work seems correct. Perhaps, with a revision of the problem, the answer was not updated.

 

[MichaelKD]

Your work seems correct. Perhaps, with a revision of the problem, the answer was not updated.

Thank you! For the next part of the problem, it asks me to find the total angular momentum of the system, is that just the angular momentum of the pulley or do I need to factor in the hanging mass as well?

 

[gneill]

Thank you! For the next part of the problem, it asks me to find the total angular momentum of the system, is that just the angular momentum of the pulley or do I need to factor in the hanging mass as well?

I suspect that they're looking for the pulley's angular momentum. But since the rotation of the pulley will increase without limit (assuming ideal components without limiting factors), you might want to express the angular momentum as a function of time.

 

[kuruman]

- ma = mg-T

How do you figure? If up is positive and down negative, the net force is
$F_{net}=T-mg$ and is a negative number because the weight must be greater than the tension for the mass to drop.
Since the counterweight is accelerating down, the other side of the equation must have a minus sign in front of it if $a$ represents the magnitude of the acceleration.
Therefore
$T-mg=-ma~$ is the correct form of the equation in which both sides are negative.

With this correction you get the given answer.

 

[haruspex]

Homework Statement: A counterweight of mass m = 4.00 kg is attached to a light cord that is wound around a pulley. The pulley is a thin hoop of radius R = 8.00 cm and mass M = 2.00 kg. The spokes have negligible mass. (a) What is the magnitude of the net torque on the system about the axle of the pulley?
Relevant Equations: tau=I*alpha=Fr, F=ma

I started by summing the forces and torques to get:
- ma = mg-T
- I*alpha=Tr

I then used a=alpha*r and I=Mr^2 to combine the equations and solved for angular acceleration equals 81.75rad/s^2. Plugging this back into a torque equation I got that the net torque is 1.04Nm. However, the problem says the answer is 3.14Nm (which is my answer times 3. Does anyone know where I went wrong?

That's a very convoluted way to solve it. All part a asks for is the torque that gravity exerts about the pulley's axle: force times perpendicular distance (giving the book answer).

 

[MichaelKD]

That's a very convoluted way to solve it. All part a asks for is the torque that the weight exerts about the pulley's axle: force times perpendicular distance (giving the book answer).

That's what I tried at first, but wouldn't that only work if the tension force is equal to the weight of the mass (mg). If that were true, the net force on the mass would be 0, meaning it's at rest. Let me know if I'm thinking about something wrong here.

 

[MichaelKD]

How do you figure? If up is positive and down negative, the net force is
$F_{net}=T-mg$ and is a negative number because the weight must be greater than the tension for the mass to drop.
Since the counterweight is accelerating down, the other side of the equation must have a minus sign in front of it if $a$ represents the magnitude of the acceleration.
Therefore
$T-mg=-ma~$ is the correct form of the equation in which both sides are negative.

With this correction you get the given answer.

Isn't T-mg=-ma equivalent to ma=mg-T (the equation I used)?

 

[kuruman]

Isn't T-mg=-ma equivalent to ma=mg-T (the equation I used)?

It is, but in #1 you posted

- ma = mg-T

which is not equivalent. I don't know what you used.

 

[MichaelKD]

It is, but in #1 you posted

which is not equivalent. I don't know what you used.

My apologies, the first dash is just connoting a list. How did you get the correct answer with ma=mg-T?

 

[kuruman]

My apologies, the first dash is just connoting a list. How did you get the correct answer with ma=mg-T?

My apologies too. I didn't get the correct answer, I actually got your answer which I thought was the book's answer.

 

[kuruman]

That's a very convoluted way to solve it. All part a asks for is the torque that gravity exerts about the pulley's axle: force times perpendicular distance (giving the book answer).

Still, part (a) should have specified that the "system" consists of the counterweight plus the pulley.

 

[Lnewqban]

... Plugging this back into a torque equation I got that the net torque is 1.04Nm. However, the problem says the answer is 3.14Nm (which is my answer times 3. Does anyone know where I went wrong?

For the system to remain balanced, a counter-moment or torque must be applied on the axle of the pulley.
The value of that torque, which direction is opposite to the one created by the counterweight of mass m, must be mgR = 3.139 N-m.

If that counterweight of mass m descends, and the pulley rotates, both in an accelerated manner, is because the value of that resisting torque is less than 3.14 N-m.

 

[MichaelKD]

For the system to remain balanced, a counter-moment or torque must be applied on the axle of the pulley.
The value of that torque, which direction is opposite to the one created by the counterweight of mass m, must be mgR = 3.139 N-m.

If that counterweight of mass m descends, and the pulley rotates, both in an accelerated manner, is because the value of that resisting torque is less than 3.14 N-m.

View attachment 324683

So when I'm solving a question like this should I assume they want the torque before it rotates at all?

 

[haruspex]

For the system to remain balanced

There is no need to assume it does.

That's what I tried at first, but wouldn't that only work if the tension force is equal to the weight of the mass (mg). If that were true, the net force on the mass would be 0, meaning it's at rest. Let me know if I'm thinking about something wrong here.

Although it is not stated, I would assume the "system" is the pulley plus the mass. The tension is an internal force, so irrelevant.
It asks for the torque exerted on the system. The external forces are the gravitational pulls on the mass and the pulley, and the force from whatever is supporting the pulley.
Of those, only the gravitational force on the mass exerts a torque about the pulley centre.
Note that the angular momentum of the system is $(I+mr^2)\omega$, where I is the MoI of the pulley about its axle. So the torque-acceleration equation is $mgr=(I+mr^2)\alpha$.

 

[Lnewqban]

After or before it rotates, the torque is transferred though a solid series of links, just like a traction force is transferred between both ends of a tensioned chain.

Even if the systems has a huge inertia (it absorbs more mechanical energy), and rotation and movement are slowly beginning to happen, the output axle or shaft will "feel" the input torque.

 

[MichaelKD]

There is no need to assume it does.

Although it is not stated, I would assume the "system" is the pulley plus the mass. The tension is an internal force, so irrelevant.
It asks for the torque exerted on the system. The external forces are the gravitational pulls on the mass and the pulley, and the force from whatever is supporting the pulley.
Of those, only the gravitational force on the mass exerts a torque about the pulley centre.

Oh okay, but why does the torque then change if you include the mass in the system or leave it out? Isn't the pulley rotating the same speed either way?

 

[kuruman]

So when I'm solving a question like this should I assume they want the torque before it rotates at all?

No, don't assume that unless explicitly stated. Here is why. In a question like this you have two moving parts. The question is unclear about what the so called system consists of. It could be (a) the pulley; (b) the counterweight only; (c) the pulley and counterweight together. All three systems experience a net torque about the pulley axis. In case (a) the net torque is what you and I calculated, $TR$; in case (b) it is $(mg-T)R$; in case (c) it is the sum of the previous two, $mgR$. You have to be a mind reader to figure out which of the three the problem is asking you to find. One should not have to solve a problem by being given the answer and then reverse-engineer the solution.

 

[MichaelKD]

No, don't assume that unless explicitly stated. Here is why. In a question like this you have two moving parts. The question is unclear about what the so called system consists of. It could be (a) the pulley; (b) the counterweight only; (c) the pulley and counterweight together. All three systems experience a net torque about the pulley axis. In case (a) the net torque is what you and I calculated, $TR$; in case (b) it is $(mg-T)R$; in case (c) it is the sum of the previous two, $mgR$. You have to be a mind reader to figure out which of the three the problem is asking you to find. One should not have to solve a problem by being given the answer and then reverse-engineer the solution.

That explanation makes so much sense, thank you!!

 

[haruspex]

No, don't assume that unless explicitly stated. Here is why. In a question like this you have two moving parts. The question is unclear about what the so called system consists of. It could be (a) the pulley; (b) the counterweight only; (c) the pulley and counterweight together. All three systems experience a net torque about the pulley axis. In case (a) the net torque is what you and I calculated, $TR$; in case (b) it is $(mg-T)R$; in case (c) it is the sum of the previous two, $mgR$. You have to be a mind reader to figure out which of the three the problem is asking you to find. One should not have to solve a problem by being given the answer and then reverse-engineer the solution.

I agree, but with only two bodies it is reasonable to guess that "system" means the combination of the two.

 

[haruspex]

After or before it rotates, the torque is transferred though a solid series of links, just like a traction force is transferred between both ends of a tensioned chain.

Even if the systems has a huge inertia (it absorbs more mechanical energy), and rotation and movement are slowly beginning to happen, the output axle or shaft will "feel" the input torque.

I have no idea what you are trying to say. The question gives no basis for assuming the system is static. Rather, the solution is as discussed in posts #15 and #18.

 

[kuruman]

$\dots~$it is reasonable to guess that "system" means the combination of the two.

haruspex:
a diviner in ancient Rome basing his predictions on inspection of the entrails of sacrificial animals

 

[haruspex]

haruspex:
a diviner in ancient Rome basing his predictions on inspection of the entrails of sacrificial animals

In my case, time-reversed: a programmer poring over the entrails of crashed software to deduce what happened in the past.