Net torque on a system with a pulley and hanging mass

No, the net torque on the system would be the torque from gravity acting on the mass (which is what you said in the next sentence).torque=m(hanging)*gravity*radius(pulley)*sin(90)?Yes!
  • #1
sunniexdayzz
13
0

Homework Statement


Part A:
A 5.73 kg mass is attached to a light cord, which is wound around a pulley. The pulley is a uniform solid cylinder of radius 9.14 cm and mass 1.26 kg. The acceleration of gravity is 9.8 m/s2 . What is the resultant net torque on the system about the center of the wheel? Answer in units of kg m2/s2.
Part B:
When the falling mass has a speed of 5.92m/s, the pulley has an angular velocity of v/r. Determine the total angular momentum of the system about the center of the wheel. Answer in units of kgm2/s.
Part C:
Using the fact that τ = dL/dt and your result from the previous part, calculate the acceleration of the falling mass. Answer in units of m/s2.
(L is the angular momentum)

Please show me the steps and explain everything!
Thank you!

Homework Equations


I=1/2mr^2
a=[tex]\alpha[/tex]*r
[tex]\sum\tau[/tex]=I[tex]\alpha[/tex]

The Attempt at a Solution


I found the sum of the forces for each body (the pulley and the hanging mass) and then i found the torque to be the radius x tension in cord. then i tried to solve for [tex]\tau[/tex] but my answer is wrong .. the final equation i got is
torque = .5m(pulley)*r(m(hanging)*g/(.5m(pulley)+m(box))

i don't even know where to start with the other two because i don't have the first part

PLEASE HELP!
 
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  • #2
sunniexdayzz said:
Please show me the steps and explain everything!
:biggrin: It's your job to do the work. We're just here to help.

The Attempt at a Solution


I found the sum of the forces for each body (the pulley and the hanging mass) and then i found the torque to be the radius x tension in cord.
The radius*tension would be the torque on the pulley, but they asked for the net torque on the system.
 
  • #3
Sorry, I didn't mean to ask that. =]

Okay, I thought there would be no torque on the hanging mass because it's not rotating? I'm assuming now that that is faulty. So how would I go about figuring out the torque on the mass? If that's even what I should be trying to do
 
  • #4
sunniexdayzz said:
Okay, I thought there would be no torque on the hanging mass because it's not rotating? I'm assuming now that that is faulty.
Yes, that's faulty because we want torques about the center of the pulley (not the center of the hanging mass).
So how would I go about figuring out the torque on the mass? If that's even what I should be trying to do
Don't think in terms of torque "on the mass". Think in terms of forces acting on the system and what torque they exert about the center of the pulley. What forces act on the system? (Do internal forces contribute to the net torque?)
 
  • #5
wouldn't the weight of the mass exert a force? so that would have something to do with the tension?
 
  • #6
sunniexdayzz said:
wouldn't the weight of the mass exert a force?
Sure, gravity exerts a force on the mass (and on the pulley).
so that would have something to do with the tension?
Why do you care about the tension?
 
  • #7
Why do you care about the tension?
Because the tension is pulling the pulley and making it rotate (at least i think so)
 
  • #8
sunniexdayzz said:
Why do you care about the tension?
Because the tension is pulling the pulley and making it rotate (at least i think so)
That's true, but it's also pulling on the mass. Does it contribute to the net torque on the system? (Consider the parenthetical question I asked at the end of post #4.)
 
  • #9
I don't know the difference between internal or external forces and which ones would contribute to the torque of the system. I don't really know what else to say?.. =[
 
  • #10
sunniexdayzz said:
I don't know the difference between internal or external forces and which ones would contribute to the torque of the system.
Don't worry about it. Do this: Identify every force acting on the system. Then figure out the torque each force exerts about the center of the pulley. Add those torques (paying attention to signs) to find the net torque.
 
  • #11
Okay .. so on the system there is gravity on the pulley, the tension pulling down on the pulley, the normal force pulling up on the pulley, the tension pulling up on the mass, and the gravity pulling down on the mass. How do i find the torque on the forces on the mass if I don't know how how far the mass is from the pulley?

I think in the end I will have that the normal force and gravity of the pulley have no torque because the point of rotation is equal the point of action, and the Tension forces will cancel each other out because they are equal and opposite so I will only be left with the gravity of the mass as acting on the system?

so would the net torque on the system be
torque=m(hanging)*gravity*radius(pulley)*sin(90)?
 
  • #12
sunniexdayzz said:
Okay .. so on the system there is gravity on the pulley, the tension pulling down on the pulley, the normal force pulling up on the pulley, the tension pulling up on the mass, and the gravity pulling down on the mass.
Good.
How do i find the torque on the forces on the mass if I don't know how how far the mass is from the pulley?
All you need to know is the radius of the pulley. (Express everything symbolically before plugging in numbers.)

I think in the end I will have that the normal force and gravity of the pulley have no torque because the point of rotation is equal the point of action, and the Tension forces will cancel each other out because they are equal and opposite so I will only be left with the gravity of the mass as acting on the system?
Yes!

so would the net torque on the system be
torque=m(hanging)*gravity*radius(pulley)*sin(90)?
Yes.
 
  • #13
Finally lol I feel so dumb sometimes when it comes to physics!

Can I check my answer? I got 5.1324756 kg m^2 / s^2
 
  • #14
sunniexdayzz said:
Can I check my answer? I got 5.1324756 kg m^2 / s^2
Looks good. But please round off to a reasonable number of significant figures.
 
  • #15
For our homework, we have to go to like 5 or 6 decimal places because it's online. Thank you sooooooo much! I will work on the other two parts and hopefully I won't need your help! Thanks again!
 

1. What is net torque and how is it calculated in a system with a pulley and hanging mass?

Net torque is the measure of the overall rotational force acting on an object or system. In a system with a pulley and hanging mass, the net torque is calculated by multiplying the force of gravity acting on the hanging mass by the distance between the pivot point (fulcrum) and the point where the force is applied (pivot arm length).

2. How does the direction of the net torque affect the motion of the system?

The direction of the net torque determines the direction of the rotation of the system. If the net torque is in a counterclockwise direction, the system will rotate counterclockwise. If the net torque is in a clockwise direction, the system will rotate clockwise.

3. What factors affect the magnitude of the net torque in a system with a pulley and hanging mass?

The magnitude of the net torque in a system with a pulley and hanging mass is affected by the weight of the hanging mass, the distance between the pivot point and the point where the force is applied, and the angle at which the force is applied. A greater weight or a longer pivot arm length will result in a larger net torque, while a smaller angle of force will result in a smaller net torque.

4. Can the net torque on a system with a pulley and hanging mass ever be zero?

Yes, the net torque on a system with a pulley and hanging mass can be zero if the system is in rotational equilibrium. This means that the net torque is balanced by an equal and opposite torque, resulting in no rotation.

5. How does friction affect the net torque in a system with a pulley and hanging mass?

Friction can decrease the net torque in a system with a pulley and hanging mass by opposing the motion of the system. This can be seen in real-life situations where a pulley system may experience resistance due to friction, resulting in a decrease in the net torque and rotational speed of the system.

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