Rotate Spool: Massless Inner Cylinder & Uniform Cylinders

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Homework Help Overview

The discussion revolves around a spool system consisting of two solid, uniform outer cylinders connected by a massless inner cylinder. The problem involves analyzing the motion and rotation of the spool as a mass is lowered, leading to questions about the direction of rotation and the effects of tension and friction.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the direction of tension and the resulting rotation of the spool, questioning why the torque from tension may lead to counterclockwise rotation despite assumptions of clockwise motion. There are discussions about the effects of friction and the conditions under which the spool rolls without slipping.

Discussion Status

Several participants have provided insights into the mechanics of the spool, with some suggesting that the problem is complex and may require detailed equations to fully understand the interactions at play. There is recognition of the need to consider both torque and friction in determining the direction of rotation, but no consensus has been reached on the final interpretation.

Contextual Notes

Participants mention the importance of additional information such as the mass of the spool and the coefficient of friction, which are currently unspecified. The assumption of rolling without slipping is also a key point of discussion.

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Homework Statement


Screen Shot 2016-12-13 at 5.22.42 PM.png

Spool made of two outer disks of radius R2 which are solid, uniform cylinders, which are connected by a massless inner cylinder of radius .

Homework Equations

The Attempt at a Solution



I just have a quick question. If the mass is going down, the tension in the string would make the spool rotate counterclockwise correct? Does that mean that the spool is rotating in a counterclockwise motion while is going to the right?
Thank you
 
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One suggestion is to work the problem with the origin at the place where the spool makes contact with the table. Then there is no torque from the frictional force. One thing you might assume is that there is sufficient friction to prevent sliding. Otherwise, it could spin the other way. Additional info could be supplied, such as the mass of the spool and the coefficient of friction, etc.
 
Right. I'm just a little confused in why the torque from the tension force is opposite to the rotation of the spool.
 
Adriano25 said:
Right. I'm just a little confused in why the torque from the tension force is opposite to the rotation of the spool.
With sufficient friction I think you will find it rotates clockwise. The problem is non-trivial though, and I would need to write out the equations to see what happens. In the case of slipping, the amount of slipping could vary considerably, depending on the frictional force. With sufficient frictional force and/or no slipping, I believe it rotates clockwise. The rolling motion can also be considered to be an instantaneous rotation about the point of contact with the table if no slipping is occurring. ## \\ ## In the case of slipping, I think you could compute the torques about the center of mass to see which way it rotates. If the frictional force is small or zero, clearly it rotates counterclockwise.
 
Thanks for your replies. Yes, I'm assuming it's rolling without slipping. I get that the spool is rotated clockwise, but when we compute the torque for tension, why is it making it rotate counterclockwise? Is it because of the way that a spool works?
 
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Adriano25 said:
Thanks for your replies. Yes, I'm assuming it's rolling without slipping. I get that the spool is rotated clockwise, but when we compute the torque for tension, why is it making it rotate counterclockwise? Is it because of the way that a spool works?
Please read the addition to my last post. Additional item is without slipping, I think you will find the torque when computed about the center of mass makes it rotate the other way because the frictional force is in the other direction and R2 is larger than R1 so it adds more clockwise torque than the inner string's counterclockwise torque. ## \\ ## There are two equations to write for the motion: The sum (or difference) of the two forces equals "ma" and the torque equation for the rotating motion. Without any slipping, these two equations are connected by the additional equation that the rotating motion equals the distance traveled. In any case, I think the string length will also enter into the solution to see how hard the hanging mass tugs depending on the acceleration of the mass, etc.
 
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Oh I get it now. Thanks!
 
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Adriano25 said:
Oh I get it now. Thanks!
Please read the additions to my very last post.
 
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Great explanation. Thanks.
 
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