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

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In summary, the problem involves a string wound around a uniform disk with a force of 50.0 N pulling on one end and a frictional torque of 3.00 Nm acting at the axle. The total torque exerted on the disk is found to be 9.5 Nm.
  • #1
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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  • #2
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.
 
  • #3
Oh yeah I just realized that. so I guess what i meant to say would the net force end up being 9.5 Nm?
 
  • #4
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.
 
  • #5
oops again...
Thanks.
 

1. What is a rotational problem?

A rotational problem is a type of physics problem that involves the rotation of an object around a fixed axis. It often requires the use of principles such as torque, angular velocity, and moment of inertia.

2. What makes a rotational problem easy?

A rotational problem is considered easy if it involves simple, idealized scenarios such as a point mass rotating around a fixed axis or a uniform object rotating about its center of mass. These problems often have straightforward solutions that can be found using basic equations.

3. How do I solve an easy rotational problem?

To solve an easy rotational problem, you will need to identify the given information, determine the unknown variables, and apply relevant equations such as Newton's laws of motion and the rotational equivalent of Newton's second law. It is also important to draw clear diagrams and use correct units in your calculations.

4. What are some common mistakes to avoid when solving easy rotational problems?

Some common mistakes to avoid when solving easy rotational problems include using the wrong equations, neglecting to consider all forces and torques acting on the object, and not using the correct units in calculations. It is also important to double-check your work and make sure your final answer has the correct units and magnitude.

5. Can you provide an example of an easy rotational problem and its solution?

Sure, here is an example: A 2 kg object is attached to a string and is rotating in a horizontal circle with a radius of 1 meter. If the string is pulled with a force of 10 N, what is the angular velocity of the object? Solution: First, we need to find the net torque acting on the object. Since the string is pulling at a 90-degree angle to the radius, the torque is equal to F*r = (10 N)(1 m) = 10 N*m. Next, we can apply the rotational equivalent of Newton's second law: τ = I*α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. Since the object is rotating at a constant speed, α = 0 and we can solve for the angular velocity using the equation ω = τ/I. The moment of inertia for a point mass rotating around a fixed axis is I = mr^2, so ω = (10 N*m)/(2 kg * (1 m)^2) = 5 rad/s.

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