Rotate Spool: Massless Inner Cylinder & Uniform Cylinders

  • Thread starter Thread starter Adriano25
  • Start date Start date
  • Tags Tags
    Rotation
AI Thread Summary
The discussion revolves around the dynamics of a spool consisting of two solid, uniform outer cylinders connected by a massless inner cylinder. Participants clarify that if the mass is descending, the spool will rotate counterclockwise due to tension, but sufficient friction can cause it to rotate clockwise instead. The confusion arises regarding the torque generated by tension and its relationship to the spool's rotation direction. It is emphasized that when rolling without slipping, the torque from friction can dominate, leading to clockwise rotation despite the tension's counterclockwise torque. The conversation highlights the importance of understanding the forces and torques involved in the motion of the spool.
Adriano25
Messages
40
Reaction score
4

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
 
Physics news on Phys.org
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?
 
  • Like
Likes Charles Link
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.
 
Last edited:
  • Like
Likes Adriano25
Oh I get it now. Thanks!
 
  • Like
Likes Charles Link
Adriano25 said:
Oh I get it now. Thanks!
Please read the additions to my very last post.
 
  • Like
Likes Adriano25
Great explanation. Thanks.
 
  • Like
Likes Charles Link
Back
Top