Rotational motion-why are torques not in opposite directions

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Homework Help Overview

The problem involves a cylinder suspended by two strings, with equal tension, rolling without slipping. The objective is to find the tension in the strings and the speed of the cylinder as it falls. The discussion centers on the relationship between linear and rotational motion, particularly the torques produced by the tensions in the strings.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations governing linear and rotational motion, questioning the reasoning behind the torque equation 2Tr = I*alpha. There is confusion about the direction of torques produced by the tensions in the strings, with some participants suggesting that the torques should cancel each other out.

Discussion Status

The discussion is ongoing, with participants exploring the implications of their diagrams and the setup of the problem. Some guidance has been offered regarding the perspective from which to analyze the torques, suggesting that both tensions contribute to torque in the same direction when viewed from the correct axis.

Contextual Notes

Participants note the potential confusion arising from the initial setup of the cylinder and the angles of the strings, which may affect the analysis of the forces and torques involved.

Epiclightning
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Homework Statement


A cylinder of mass m is suspended through 2 strings wrapped around it at its ends, connected to the ceiling. Both strings have equal tension, and the cylinder rolls without slipping. r is the distance between the CM of the cylinder and each end. I is the moment of inertia of the cylinder.
Find the tension T in the string and the speed of the cylinder as it falls through a distance h.

Homework Equations


Equations for linear motion in vertical direction for center of mass of the cylinder, and that of rotational motion about the CM.

The Attempt at a Solution


Linear:
mg - 2T = ma

Rotational:
The solution says 2Tr = I*alpha (where alpha is angular acceleration).
After this we simplify using angular acceleration= a/r, and solve the equations.

I understand that part, but I don't get why 2Tr = Ia...both tensions are acting on the top of the cylinder, so the torques should be in opposite directions and cancel each other out, right? When I use torque = r x T, the directions come out of the page and into the page. But the equations say that both tensions contribute an equal torque in one direction. Why?
 
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Is their a picture or other diagram which is furnished describing how the cylinder is suspended initially?
 
image.jpg
 
Epiclightning said:

Based on this sketch, it appears that the cylinder is initially suspended with both strings coming off one side.

You should sketch a free body diagram of the cylinder and the strings to determine the total forces acting on the cylinder as it starts to move.
 
I did, and that's how I got the linear equations, but the torques appear to be in opposite directions, unlike the equation that the solution suggests (2Tr = I*alpha). This equation suggests that the torques are in the same direction.
 
Epiclightning said:
I don't get why 2Tr = Ia...both tensions are acting on the top of the cylinder, so the torques should be in opposite directions and cancel each other out, right? When I use torque = r x T, the directions come out of the page and into the page.
That's because you are looking at torque around the wrong axis. The diagram is confusing you. Consider what it looks like from one end of the cylinder. You should see a circle (cylinder end) with a string rising almost vertically from one side. The other string is directly behind it. Each string exerts the same torque about the cylinder's axis.
Note that it's really a bit more complicated. The strings won't quite be vertical, but you can't take that into account because it depends on string length.
 
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Ohh, yes that makes sense. Then both torques would be in the same direction. Thanks :)
 

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