Angular acceleration caused by mass over a pulley

[Fluxthroughme]

The method I'm using in the image is the one that makes the most sense to me, but I've tried others and still cannot get at the stated answer, which is 2.65 rad/s^2

 

[Doc Al]

Do not assume that the tension in the rope equals the weight of the hanging mass. Set up two equations, one for the post and one for the hanging mass. Then you can solve for the acceleration.

 

[Fluxthroughme]

Do not assume that the tension in the rope equals the weight of the hanging mass. Set up two equations, one for the post and one for the hanging mass. Then you can solve for the acceleration.

Sorry, how do I go about that? If the tension in the string isn't 5g, I don't know how to do it.

 

[Doc Al]

Sorry, how do I go about that? If the tension in the string isn't 5g, I don't know how to do it.

Don't guess, solve! Call the tension T and set up your equations. That will be one of the unknowns. Luckily, you'll have two equations and two unknowns.

 

[Fluxthroughme]

Don't guess, solve! Call the tension T and set up your equations. That will be one of the unknowns. Luckily, you'll have two equations and two unknowns.

I'm not even guessing; I just don't know how to set up the equations. T - 5g = 5a, I think? Then I am 100% lost at the post.

 

[Doc Al]

I'm not even guessing; I just don't know how to set up the equations. T - 5g = 5a, I think?

Good! Since the mass falls, I would choose positive as down and let "a" stand for the magnitude of the acceleration.

Now write an equation for the post.

 

[Fluxthroughme]

Good! Since the mass falls, I would choose positive as down and let "a" stand for the magnitude of the acceleration.

Now write an equation for the post.

I'm not sure how I can be any more direct: I am 100% lost at the post. It's not for lack of thinking, I just don't know what the equation ought to be?

 

[Doc Al]

I'm not sure how I can be any more direct: I am 100% lost at the post. It's not for lack of thinking, I just don't know what the equation ought to be?

You already set up the equation for the post in your first post! Your mistake was assuming that the tension was equal to mg. Set the unknown tension equal to T and write out that equation.

 

[Fluxthroughme]

You already set up the equation for the post in your first post! Your mistake was assuming that the tension was equal to mg. Set the unknown tension equal to T and write out that equation.

If you mean [itex]rF=I/alpha[/itex], then isn't that a different acceleration to the acceleration for the block?

 

[Doc Al]

If you mean [itex]rF=I/alpha[/itex], then isn't that a different acceleration to the acceleration for the block?

Yes, that's what I mean. Almost. It should be rF = I*alpha.

Yes, one is an angular acceleration and the other is a linear acceleration. How are they related? (The rope is attached to the post.)

 

[Fluxthroughme]

Yes, that's what I mean. Almost. It should be rF = I*alpha.

Yes, one is an angular acceleration and the other is a linear acceleration. How are they related? (The rope is attached to the post.)

Yeah, that's what I meant - LaTeX troubles :P

Well, regardless, I still get the wrong answer:

And I checked, in case I mixed my signs up, adding 6.25 rather than subtracting, which gives 2.84, still not the correct answer.

 

[Doc Al]

Well, regardless, I still get the wrong answer:

You need to fix the sign of "a" in your equation for the hanging mass.

And how does alpha relate to "a"?

 

[Fluxthroughme]

You need to fix the sign of "a" in your equation for the hanging mass.

Which doing will give me 2.84 as the final answer, which is still incorrect? =/

 

[Doc Al]

Which doing will give me 2.84 as the final answer, which is still incorrect? =/

Write your equations. Tell me how alpha relates to "a".

Don't solve them, just write them.

(I solved it and got the correct answer, so we need to see where you are going wrong.)

 

[Fluxthroughme]

Write your equations. Tell me how alpha relates to "a".

Don't solve them, just write them.

(I solved it and got the correct answer, so we need to see where you are going wrong.)

I don't know how it relates it general, but at that instant, they are the same?

5g - T = 5a
[itex]\tau = rF=r(5g-5a) = 61.3 - 6.25a = Ia, a(I+6.25) = 61.3[/itex]
[itex]\frac{61.3}{\frac{1}{3}*15*1.75^2 + 6.25} = a[/itex]
a = 2.84

 

[Doc Al]

I don't know how it relates it general, but at that instant, they are the same?

How can an angular acceleration (in rad/s^2) be the same as a linear acceleration (in m/s^2)?

But they are related. Consider the point of attachment of the rope with the post. How does the linear acceleration of that point relate to the angular acceleration of the post?

You're almost there.

 

[Fluxthroughme]

How can an angular acceleration (in rad/s^2) be the same as a linear acceleration (in m/s^2)?

But they are related. Consider the point of attachment of the rope with the post. How does the linear acceleration of that point relate to the angular acceleration of the post?

You're almost there.

Ok, [itex]a=r\alpha[/itex]

Solving from there is just basic algebra. Prior to googling, I had not seen this, nor thought about a concrete relationship between them, which must've been why I could not see how to do the problem. Thank you for putting up with me :P

 

[Doc Al]

Ok, [itex]a=r\alpha[/itex]

Solving from there is just basic algebra. Prior to googling, I had not seen this, nor thought about a concrete relationship between them, which must've been why I could not see how to do the problem. Thank you for putting up with me :P

Now you've got it. You should be able to the the correct answer now.

And you're welcome.