# Mechanics Problem

Discussion in 'Homework Help' started by lendo1, May 8, 2011.

1. ### lendo1 Thread Starter Active Member

Apr 24, 2010
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Say you have a pulley, with radius, R, and mass, M.

This pulley is a disc with rotational inertia given by: I = (1/2)MR^2

This pulley has a string wrapped around it with negligible friction, so that the pulley can be dropped, with the end of the string attached to a fixed point. The pulley, thus, appears to roll down the string.

The pulley is dropped. What is the acceleration of the pulley in terms of the given variables and fundamental constants?

Feb 5, 2010
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3. ### lendo1 Thread Starter Active Member

Apr 24, 2010
34
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Very good Kermit, this is on Earth. So you have two forces acting on the pulley, the force of gravity (M*g), and the upward pull of the string. What is the acceleration?

4. ### Georacer Moderator

Nov 25, 2009
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Wouldn't it be nice if you had posted that image instead of me?

Try this set of equation for starters:
$\left{ \begin{array}{c} \Sigma F = M \cdot \alpha \\
\Sigma M = I \cdot \gamma \\
\alpha = \gamma \cdot R \end{array} \right$

Where $\alpha$ is the linear acceleration,
$\gamma$ is the rotational acceleration and
$M$ is the cylinder's torque.

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5. ### lendo1 Thread Starter Active Member

Apr 24, 2010
34
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That's it, thanks for the drawing! I tried using those equations (and others) and every time I have ended up setting torque to torque [(Rotational Inertia) * (Angular acceleration)] = (Tension * Radius), and my answers never seem to make any sense... too many variables cancel out; there is a property I am forgetting or an equation I am misusing, but I can't figure out where!

6. ### Georacer Moderator

Nov 25, 2009
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1,266
Care to write the set of equations I gave you down and work with them a bit? The Homework Help section requires you to do some work too, as you might remember.

Can you find an expression for the sum of the forces (as a function of the forces I noted on the sketch?)
Can you find an expression for the sum of the torques, with a reference to the center of the cylinder?

If you do that, using the proposed set of equations, you will end up with three equations and three unknowns.

7. ### lendo1 Thread Starter Active Member

Apr 24, 2010
34
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Let me show you what I came up with, please point me in the right direction

My variables: T-tension, g-gravitational acc, a - linear acceleration, alpha - angular acceleration, R- radius, M - mass, I - Rotational Inertia

Sum of forces:
T - M*g = -M*a

Sum of torques:
TorqueNet = ?

Isn't there only one torque acting on the disc? Torque = T * R ? Which has to be equal to (I * alpha)? That is the loop I keep finding myself in. I don't see how I can make a reference to the center of the cylinder using anything other than R, but that puts me back into this same situation where I*alpha = T*R, (which leaves me with two unknowns, tension AND linear acceleration).

I ended up with a = (2T)/M
Substituting that into my sum of forces equation gives me T = -Mg.... which can't be right

8. ### Georacer Moderator

Nov 25, 2009
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Ok, another info income shot:
$\left{ \begin{array}{l}
Weight-Tension=M \cdot \alpha \\
Tension \cdot R=I \cdot \gamma\\
\alpha=R \cdot \gamma \end{array}$

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9. ### lendo1 Thread Starter Active Member

Apr 24, 2010
34
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Well I tried substituting into (Tension*R = I*/gamma) and I ended up with a = (2/3)*g. Definitely a believable result, but that means acceleration is independent of mass and radius... which I suppose makes sense. Is this right?

Thanks!

10. ### Kermit2 AAC Fanatic!

Feb 5, 2010
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UPWARD pull of a string?

You need to word your questions more carefully.

Quote: "The pulley is dropped. What is the acceleration....?"

You DID NOT give the pulley any definite rotation to start with, but simply defined a formula giving rotational inertia. Within the confines of your question my answer was totally appropriate----whether you find it helpful or not, would be up to you.

Kermit,
Earth

11. ### lendo1 Thread Starter Active Member

Apr 24, 2010
34
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Yes, the UPWARD pull of the string. Relative to the disc, the force of tension from the string is UP, opposing the downward force of gravity. It was a tricky problem, Kermit, no worries.

12. ### Georacer Moderator

Nov 25, 2009
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1,266
What Kermit is trying to teach you is to pick your words more carefully.
The string isn't moved upwards and therefore the word "pull" isn't suitable. A better phrase would be "and the upwards tension from the string".

Feb 5, 2010
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