Find the force as a function of radius to pull a ribbon

Click For Summary

Homework Help Overview

The problem involves a disc that rotates about its center, with a ribbon wrapped around it. The task is to determine the force required to pull the ribbon as a function of the radius of the disc, considering the dynamics of the system and potential frictional effects.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of friction on the problem and whether it should be considered. There is debate about the relationship between linear and angular velocities as the radius changes. Some participants express confusion about how to relate force to the changing radius and the resulting angular acceleration.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have suggested starting points for analysis, while others are seeking clarification on the assumptions regarding angular acceleration and the relationship between the variables involved.

Contextual Notes

There is uncertainty regarding the presence of friction and how it affects the dynamics of the disc and ribbon system. Participants are also considering the implications of a changing radius on the angular acceleration of the disc.

Kanda ryu
Messages
14
Reaction score
1

Homework Statement



A disc is free to rotate about an axis passing through its center and perpendicular to its plane. The moment of inertia of the disc about its center is I. A light ribbon is tightly wrapped over it in multiple layers. The end of the ribbon is pulled out at a constant velocity 'u'. Let the radius of the ribboned disc be 'R' at any time and thickness of ribbon be 'd' (d<<R).Find the force (F) required to pull the ribbon as a function of radius R.

Homework Equations

The Attempt at a Solution



What confuses me is whether friction is present or not. The question doesn't clearly specify about friction. Assuming friction exists, the question gets difficult to solve. As it is being pulled with uniform velocity, net torque about the center of disc should be zero, so I tried equating net torque at R and and a small change in radius R-dx but it doesn't seem to help .I need a little hint on understanding the question and how to start off with it.
 
  • Like
Likes   Reactions: ZeroFive
Physics news on Phys.org
Kanda ryu said:
As it is being pulled with uniform velocity, net torque about the center of disc should be zero
The radius of disc+layers of ribbon does decrease, albeit slowly. So if the ribbon moves with constant u, the disc still needs some angular aceleration.
 
So equating torque is completely out of use as angular acceleration would be different as R decreases right? But isn't the ribbon pulled in a way that the ribbon has a constant linear velocity so the ribbon in contact with disc will also have constant velocity. Assuming no slipping between ribbon and disc, disc should also have constant tangential velocity, hence constant angular velocity. If not then how do I get a relation between force and change in radius?
Here is the diagram.
 

Attachments

  • IMG_20180326_163748.jpg
    IMG_20180326_163748.jpg
    20.1 KB · Views: 508
Kanda ryu said:
constant tangential velocity, hence constant angular velocity.
With a changing radius??
 
Alright I get it, there is angular acceleration. Could you give me a hint on starting the question?
 
Assume some ##\omega## and ##R## at ##t=0## (\ for example ##\omega_0## and ##R_0##\ ) and start applying your relevant equations ( oops...:rolleyes: )
 
Last edited:
Kanda ryu said:
Alright I get it, there is angular acceleration. Could you give me a hint on starting the question?
The usual approach: define some variables that will be functions of time and write some equations relating them to each other and to the given constants.
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
2K
Replies
43
Views
3K
Replies
1
Views
2K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
7
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
9
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K