Calculating Work Done on Pulley by Falling Block

Since the pulley is a uniform disk, its moment of inertia is 1/2*mass*radius^2 = 1/2*2.4(.220)^2 = 0.02856 kgm^2. So the torque is 0.02856*0.180 = 0.00514 Nm. And the work done by the block on the pulley is torque*angular displacement = 0.00514*2.27 = 0.01168 J. In summary, as the block falls 0.500 m, it does 0.01168 J of work on the pulley, which is a uniform disk with a mass of 2.40 kg
  • #1
r_swayze
66
0
The pulley in the illustration is a uniform disk of mass 2.40 kg and radius 0.220 m. The block applies a contant torque to the pulley, which is free to rotate without friction, resulting in an angular acceleration of magnitude 0.180 rad/s2 for the pulley. As the block falls 0.500 m, how much work does it do on the pulley?

The illustration is of a pulley with a rope hanging a block down vertically.

Here is my attempt:

work = torque x angular diplacement

If the block falls then the pulley turns by the same amount, right?

so arc length = .500m

theta = arc length / radius = .500 / .220 = 2.27 radians = angular displacement?? (is this right?)

To find torque, I need F x r

r = .220

How do I find the Force?

Do I use the mass = 2.4 kg and multiply by tangential acceleration?
 
Physics news on Phys.org
  • #2
Yes. It is tangential acceleration.
How to find out the tangential acceleration?
 
  • #3
tangential acceleration is angular acceleration times radius

but what about angular displacement? Did I do that right?
 
  • #4
Your angular displacement is correct.
Now the torque = moment of inertia*angular acceleration.
 

1. How do you calculate work done on a pulley by a falling block?

To calculate the work done on a pulley by a falling block, you need to know the displacement of the block, the force applied to the block, and the angle between the force and the displacement. The formula for work is W = Fdcosθ, where W is the work done, F is the force, d is the displacement, and θ is the angle between the force and the displacement.

2. What is the role of a pulley in calculating work done by a falling block?

A pulley is used to change the direction of the force applied to the block. It also helps to distribute the weight of the block evenly, making it easier to calculate the work done. Without a pulley, the force would need to be applied directly on the block in the direction of its displacement.

3. How does the mass of the block affect the work done on the pulley?

The mass of the block does not directly affect the work done on the pulley. However, it does affect the force required to lift or lower the block, which in turn affects the work done. The heavier the block, the more force is needed to move it, and therefore more work is done on the pulley.

4. Can the work done on a pulley by a falling block be negative?

Yes, the work done on a pulley can be negative if the force applied to the block is in the opposite direction of its displacement. This means that the pulley is actually doing work on the block, rather than the block doing work on the pulley.

5. How is the work done on a pulley related to the tension in the rope?

The work done on a pulley is directly related to the tension in the rope. The tension in the rope is equal to the force applied to the block, and as mentioned before, the force is a crucial factor in calculating work. Therefore, the greater the tension in the rope, the greater the work done on the pulley.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
417
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
30
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
977
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top