Calculating Work Done on Pulley by Falling Block

  • Thread starter Thread starter r_swayze
  • Start date Start date
  • Tags Tags
    Rotational Work
AI Thread Summary
The discussion focuses on calculating the work done on a pulley by a falling block. The pulley is a uniform disk with a mass of 2.40 kg and a radius of 0.220 m, experiencing an angular acceleration of 0.180 rad/s². As the block falls 0.500 m, the arc length corresponds to the angular displacement of the pulley, calculated to be approximately 2.27 radians. To find the torque, the force exerted by the falling block is determined using its mass and tangential acceleration, which is derived from the angular acceleration multiplied by the radius. The final calculation for torque involves the moment of inertia and angular acceleration, confirming the relationship between linear and angular motion.
r_swayze
Messages
65
Reaction score
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
Yes. It is tangential acceleration.
How to find out the tangential acceleration?
 
tangential acceleration is angular acceleration times radius

but what about angular displacement? Did I do that right?
 
Your angular displacement is correct.
Now the torque = moment of inertia*angular acceleration.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top