Physics Help Forum Rotating and translation on a frictionless surface

 Jan 12th 2017, 06:40 AM #1 Junior Member   Join Date: Jan 2017 Posts: 3 Rotating and translation on a frictionless surface A uniform spool of mass M and diameter d rests on end on a frictionless table. A massless string wrapped around the spool is attached to a weight m which hangs over the edge of the table. If the spool is released from rest when its center of mass is a distance l from the edge of the table, what is the velocity of the weight m when the center of mass of the spool reaches the edge of the table? My attempt: I thought of breaking up the problem into two cases and the combining them at the end. case1: Pretend no rotation: With no rotation the spool has forces Tension acting on it. $\displaystyle T = Ma$ The mass attached to the string has forces Tension and gravity. solved for $\displaystyle T' = mg - ma$ Since the acceleration for both we can get to $\displaystyle a = (mg)/(M+m)$ So, we can get a final velocity of $\displaystyle v = (2*(mg)/(M+m)*)^1/2$ where I started with $\displaystyle vf^2 = vi^2+2al$, l being the displacement of the spool on the table. Case2: Pretend no translation: With no translation, I believe then the Tension and Torque are equal to eachother. Then we can get $\displaystyle alpha = (torque/I)$. and we can get $\displaystyle theta = L/(pi*d)$, (note: used 'L' instead of lowercase 'l' to see clear difference between the I, moment of inertia.). What I end up using is the angular kinematics to get $\displaystyle omegaf= (2*(torque/I)*(L/p(i*d)))^1/2$ So this is my work...am I on the rigth track or completely wrong? And how can I relate these two to get a uniform equation?
 Jan 12th 2017, 10:13 AM #2 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,240 You can't separate translation from rotation. The issue is that the linear acceleration of the weight is nit the same as the linear acceleration of the spool - as the spool unwinds the weight drops faster than the spool moves across the table. I suggest starting with the relationship between displacement of the weight and of the spool. If we let y = the displacement of the weight and x = displacement of the spool, then: $\displaystyle y = x + r \theta$ and $\displaystyle \ddot y = r \ddot \theta$ The sum of forces acting on the weight of mass m: $\displaystyle W-T = m\ddot y = m(\ddot x + r \ddot \theta )$ where W is weight of the mass m and T is the tension in the string. The sum of forces acting on the spool of mass M: $\displaystyle T = M\ddot x$ Sub this expression for T back into the previous equation, and you can solve for $\displaystyle \ddot x$. Can you take it from here?
Jan 12th 2017, 11:53 AM   #3
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 Originally Posted by ChipB You can't separate translation from rotation. The issue is that the linear acceleration of the weight is nit the same as the linear acceleration of the spool - as the spool unwinds the weight drops faster than the spool moves across the table. I suggest starting with the relationship between displacement of the weight and of the spool. If we let y = the displacement of the weight and x = displacement of the spool, then: $\displaystyle y = x + r \ theta$ and $\displaystyle \ ddot y = r \ ddot \ theta$ The sum of forces acting on the weight of mass m: $\displaystyle W-T = m\ddot y = m(\ddot x + r \ddot \theta )$ where W is weight of the mass m and T is the tension in the string. The sum of forces acting on the spool of mass M: $\displaystyle T = M\ddot x$ Sub this expression for T back into the previous equation, and you can solve for $\displaystyle \ddot x$. Can you take it from here?
So my related equation would be $\displaystyle mg - M\ddot x = m(\ddot x +r \ddot \theta )$?
solving for the accleration of x would give me the accerlation of the spool, but I need the velocity of the weight. I think my source of confusion is the $\displaystyle W - T = m \ddot y = m( \ddot x + r \ddot \theta)$ Are you saying the m \ddot y is equal to $\displaystyle m( \ddot x + r \ddot \theta)$? Or the whole thing? I understand that m( \ddot x + r \ddot \theta) is the total force of the spool contributing to the weight, but how then will I get a final velocity for the weight if its all the x accleration?

Last edited by hadsox; Jan 12th 2017 at 01:23 PM.

 Jan 12th 2017, 01:14 PM #4 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,240 I guess there's a bit more to go here... From $\displaystyle T = m\ddot x = \frac {I \ddot \theta}r$ you can relate $\displaystyle \dot \theta$ to $\displaystyle \dot x$ , and hence to $\displaystyle \dot y$. This should enable you to relate total vertical displacement of the weight (call it h) to the horizontal displacement of the spool, L. I think then using energy principles would bring it home: $\displaystyle KE_{spool} + KE_{weight} = mgh$
Jan 12th 2017, 01:43 PM   #5
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 Originally Posted by ChipB I guess there's a bit more to go here... From $\displaystyle T = m\ddot x = \frac {I \ddot \theta}r$ you can relate $\displaystyle \dot \theta$ to $\displaystyle \dot x$ , and hence to $\displaystyle \dot y$. This should enable you to relate total vertical displacement of the weight (call it h) to the horizontal displacement of the spool, L. I think then using energy principles would bring it home: $\displaystyle KE_{spool} + KE_{weight} = mgh$
I've worked the problem to have:

$\displaystyle mg - \frac {I \ ddot \theta}r = m( \ ddot x + r \ ddot \theta)$

I was given what I is equal to, so I can plug that in getting:
$\displaystyle mg - \frac {Md^2 \ddot \theta}8}r = m( \ ddot x + r \ ddot \theta)$
I'm still not sure how to go back to velocity of the weight though, How can I solve for \ddot x and have that be the acceleration of the falling weight?

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