Atwoods: Deriving Torque Relation with R

  • Context: Undergrad 
  • Thread starter Thread starter breez
  • Start date Start date
  • Tags Tags
    Torque
Click For Summary

Discussion Overview

The discussion revolves around deriving the torque relation for a pulley system with two masses, focusing on the role of forces and distances involved in calculating net torque. The scope includes theoretical reasoning and mathematical derivation related to mechanics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that the total external torque about the pulley center is given by (m1 – m2)gR and asks for clarification on the derivation and the significance of the distance R.
  • Another participant explains that the force mg acts vertically and that R is the perpendicular distance to the axis, leading to the torque expression τ = RF = mgR.
  • Some participants argue that the tension, rather than gravity, is the force acting perpendicular to the center of the pulley, raising questions about the net torque calculation.
  • A participant expresses confusion about calculating net torque, suggesting that individual torques should be added and questioning the reasoning behind using R as a distance.
  • Another participant clarifies that while the distance r is changing, the relevant distance for torque calculation is r sin θ, which remains constant and equals R.
  • A later reply acknowledges the importance of the moment arm in torque calculations, indicating a realization of the cross product's role.

Areas of Agreement / Disagreement

Participants express differing views on the forces contributing to torque and the appropriate method for calculating net torque. There is no consensus on the best approach to derive the torque relation, and the discussion remains unresolved.

Contextual Notes

Participants mention the need to consider the center of mass of the attached masses and the changing distances involved, indicating potential limitations in their reasoning and assumptions about the system.

breez
Messages
65
Reaction score
0
A pulley with radius R is free to rotate on a horizontal fixed axis through its center. A string passes over the pulley. Mass m1 is attached to one end and mass m2 is attached to the other. The portion of the string attached to m1 has tension T1 and the portion attached to m2 has tension T2. The magnitude of the total external torque, about the pulley center, acting on the masses and pulley, considered as a system, is given by:

(m1 – m2)gR

Can someone explain how to derive this relation? Why is the distance just R?
 
Physics news on Phys.org
The force (mg) acts vertically and R is the perpendicular distance to the axis. So each torque is just:

[tex]\tau = \vec{r} \times \vec{F} = RF = mgR[/tex]
 
I thought the force a acting perpendicular to the center of pulley is tension, not gravity.
 
breez said:
I thought the force a acting perpendicular to the center of pulley is tension, not gravity.
The force on the pulley is the tension, not the weight. But I thought you were asking about the net torque on the entire system, not just on the pulley. (A perfectly valid way to analyze the problem, but I would not recommend it.)
 
Hmm,

The problem asks for the net torque on the entire system about the center of the pulley. I believe the gravitational forces, m1g and m2g are acting at the center of masses of the two masses, which are some distance from the center. I am confused how to calculate the net torque because I thought you had to add all the individual torques, which would mean I would need to know the distance at any time between the center of mass of each mass and the center of the pulley.

I'm sorry, but I do not see the reasoning behind just adding up all the forces on the system and multiplying by R.
 
breez said:
The problem asks for the net torque on the entire system about the center of the pulley. I believe the gravitational forces, m1g and m2g are acting at the center of masses of the two masses, which are some distance from the center.
You can certainly do it that way also. You'll get the same answer.
I am confused how to calculate the net torque because I thought you had to add all the individual torques, which would mean I would need to know the distance at any time between the center of mass of each mass and the center of the pulley.
All you need is the perpendicular distance. Consider this:
[tex]\tau = \vec{r} \times \vec{F} = rF\sin\theta[/tex]

While the distance r is constantly changing, what counts is [itex]r\sin\theta[/itex]--which is constant and equal to R.
I'm sorry, but I do not see the reasoning behind just adding up all the forces on the system and multiplying by R.
You need to firm up your understanding of torque and how to calculate it.
 
Yeah, you're right. I forgot it was the moment arm, since it's the cross product. Wow.
 

Similar threads

Undergrad Angular acceleration of a massless pulley
  • · Replies 8 ·
Replies
8
Views
7K
Graduate Does the Relation ##T_2 = T_1 e^{\mu \theta}## Hold for an Accelerating Capstan?
  • · Replies 1 ·
Replies
1
Views
1K
Calculating tensions and acceleration
  • · Replies 10 ·
Replies
10
Views
3K
Undergrad Moment of inertia of human body with limbs at an angle
  • · Replies 12 ·
Replies
12
Views
2K
Undergrad Properties of angular momentum
  • · Replies 6 ·
Replies
6
Views
1K
Undergrad Acceleration on an electric unicycle, how much does the rider have to lean?
  • · Replies 30 ·
2
Replies
30
Views
4K
Torque about an accelerating point
  • · Replies 5 ·
Replies
5
Views
3K
High School Sign conventions in torque and non-uniform circular motion
  • · Replies 1 ·
Replies
1
Views
2K
Pulleys with Torque: Free Body Diagram Analysis
  • · Replies 34 ·
2
Replies
34
Views
4K
Why are the forces on torque tension and not the weight of the mass?
  • · Replies 7 ·
Replies
7
Views
2K