How do I calculate the total torque exerted on a rotating disk?

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Homework Help Overview

The discussion revolves around calculating the total torque exerted on a rotating disk, which is influenced by a pulling force and a frictional torque. The problem involves a uniform disk with specific dimensions and mass, and participants are exploring the implications of the given forces and torques.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of calculating net torque by considering both the applied force and the frictional torque. There is a question regarding whether the radius should be used for the frictional torque, as it is provided directly as a torque value.

Discussion Status

The discussion is active, with participants clarifying their understanding of how to incorporate the frictional torque into their calculations. Some have suggested that the net torque can be determined without using the radius for the frictional component, leading to a reevaluation of the calculations presented.

Contextual Notes

There is a focus on ensuring the correct interpretation of the problem's parameters, particularly regarding the frictional torque and its application in the overall torque calculation. Participants are navigating through potential misunderstandings related to the definitions of torque and force in this context.

Sheneron
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[SOLVED] Easy rotational problem

Homework Statement


A string is wound around a uniform disk of radius 0.25m and mass 8.00kg. The end of the string is pulled with a constant force of 50.0 N, and there is a frictional torque acting at the axle of 3.00 Nm.
A) Find the total torque exerted on the disk

The Attempt at a Solution


I think I have it solved... just wanted to make sure and I have one question. It says that the frictional torque is acting on the axle, do I still use the radius r=0.25 to find the torque for that part? Here is what I did:

[tex]\Sigma \tau = \Sigma Fd[/tex]
[tex]\Sigma \tau = (50)(0.25) + (-3)(0.25)[/tex]
[tex]\Sigma \tau = 11.75 Nm[/tex]

So is that how I would find the net torque?
 
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Sheneron said:
It says that the frictional torque is acting on the axle, do I still use the radius r=0.25 to find the torque for that part?
Careful. For friction they gave you the torque, not the force. So you don't have to use any radius to find the frictional torque.
 
Oh yeah I just realized that. so I guess what i meant to say would the net force end up being 9.5 Nm?
 
Sheneron said:
so I guess what i meant to say would the net force end up being 9.5 Nm?
The net torque is 9.5 Nm.
 
oops again...
Thanks.
 

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