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jgrins jgrins
wrote...
Posts: 15
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7 years ago
A frictionless pulley, which can be modeled as a 0.86kg solid cylinder with a 0.31m radius, has a rope going over it, as shown in the figure. (Figure 1 attached)
The tensions in the rope are 12N and 10N . What is the angular acceleration of the pulley?

I've attached the drawing I did so far for the problem. I know I need to use either the moment arm or the radial arm but I don't know the angles to find them, so I'm not sure what the next step is.

I know I need to find the net torque that is exerted by the pulley, and I can find the moment of inertia by I=1/2MR2.

If someone could give me a hint, that would be great, not necessarily to solve. Thanks!

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wrote...
7 years ago
A frictionless pulley, which can be modeled as a 0.86kg solid cylinder with a 0.37m radius, has a rope going over it, as shown in the figure.

The torque is (12 - 10)*0.37 N-m = 0.74 N-m
The moment of inertia of a solid cylinder is ½*m*r² = 0.05887 kg-m²

The angular acceleration is T/I = 12.57 kg-m² which is the correct answer except that it should have only 2 significant figs;

a = 13 kg-m²
Biology - The only science where multiplication and division mean the same thing.
wrote...
7 years ago Edited: 7 years ago, jgrins
Okay but how did you find torque = (12-10)*.037 N-m?
I know that you subtract the 10N force bc it's clockwise, but where did you get the value .37 N-m? Is it the same as the radius?

Sorry, I want to understand as well, not just get an answer.
Post Merge: 7 years ago

And are you assuming that the 12-10N is the perpendicular force, because it looks like they're at an angle with the figure?
wrote...
7 years ago
Torque = T = (12-10) * 0.3  = 0.6

I = 0.5 m*r2 = 0.03645

So α = T/I

α = 16.46 rad/s2
Answer accepted by topic starter
bio_manbio_man
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Educator
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7 years ago
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Guys, if you're going to use examples with different values, please make this values known.

Try this:

A frictionless pulley which can be modeled as a 0.80 solid cylinder with a 0.30 m radius has a rope going over it. The tension in the rope is 10 N on the right side and 12 N on the left side. What is the angular acceleration of the pulley?

Torque= Moment of Intertia x angular acceleration

torque= Fxr = (12-10N) x (0.35m) = 0.70 N*m

The moment of intertia for a solid cylinder is 1/2 MR^2, or 1/2 (0.87kg)*(0.35)^2 = 0.0532 kg*m^2

0.70 = 0.0532 * angular acceleration

Acceleration = 0.70/0.0532 = 13.158 rad/s^2

OR
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wrote...
7 years ago
Did any of these help you?
Biology - The only science where multiplication and division mean the same thing.
wrote...
7 years ago
Yeah they did. I messed up because I assumed there was an angle between the tension and the pulley but my professor told me its just perpendicular to the pulley so torque=rF
wrote...
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7 years ago
Awesome, marking it solved Slight Smile
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