Work Done by String Connecting to Uniform Vertical Disk | Solution and Equations

Click For Summary

Homework Help Overview

The problem involves a light string connected to a uniform vertical disk, which is free to rotate about a horizontal axis. The disk starts at rest, and a constant horizontal force is applied to the string as the disk makes a quarter revolution. The task is to determine the work done by the string during this process.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of work done and torque, with initial attempts involving integrals and expressions for torque. Questions arise regarding the correct angle for a quarter revolution and the nature of the force application.

Discussion Status

There is an ongoing exploration of the problem's setup, particularly regarding the relationship between the applied force and the rotation of the disk. Some participants express confusion over the assumptions made about the force's application point and its effect on torque calculations. Clarifications are being sought, and some guidance has been provided regarding the interpretation of the problem.

Contextual Notes

No visual aids are provided with the problem, and participants note the importance of understanding the force's application in relation to the disk's rotation. The original poster emphasizes that the force is constant and horizontal, which influences the discussion about the torque and work done.

Amar.alchemy
Messages
78
Reaction score
0
Work done by String??

Homework Statement


You connect a light string to a point on the edge of a uniform vertical disk with radius R and mass M. The disk is free to rotate without friction about a stationary horizontal axis through its cente. Initially, the disk is at rest with the string connection at the
highest point on the disk. You pull the string with a constant horizontal force F until the wheel has made exactly one-quarter revolution about a horizontal axis through its center, and then you let go. (a) Find the work done by the string


Homework Equations




The Attempt at a Solution


[tex] \[\begin{array}{l}<br /> w = \tau \int\limits_0^{\frac{\pi }{4}} {d\theta } \\ <br /> w = FR\pi /4 \\ <br /> \end{array}\][/tex]

Is this correct??Kindly help me
 
Physics news on Phys.org


It is correct with one exception. What fraction of π is one-quarter of a revolution?
 


Sorry, it shuld be [tex]\[\frac{\pi }{2}\][/tex]...

However in the solution manual it is given as:
The torque is [tex]\[\begin{array}{l}<br /> \tau = FR\cos \theta \\ <br /> w = \int\limits_0^{\frac{\pi }{2}} {FR\cos \theta d\theta } \\ <br /> w = FR \\ <br /> \end{array}\][/tex]

So i am little bit confused... :confused:
 


Is there a picture that goes with this? Your answer would be correct if the tension pulling on the wheel remained tangent to the wheel at all times as in a pulley. This is what I thought initially to be the case. Here it seems that the string is firmly attached to the rim of the wheel and the point of application of the force rotates with the wheel.

Draw yourself a wheel and put in a horizontal tension F at some angle, say the "1 o'clock" position. Call θ the angle between 12 o'clock and 1 o'clock. What is the torque? Note that the lever arm changes from R at 12 o'clock to zero at 3 o'clock.
 


There is no picture given with this question and in the problem he clearly mentions that "The force applied is constant and horizontal"... so do you think is it appropriate to assume that point of application of the force rotates with the wheel??
 


Yes, I think so. As you say, "clearly mentions it." I misread the problem initially. Also you get the answer in the solution manual that way.
 


ok, now i understood the problem (the string is connected at a point instead of wrapping around the wheel)... Thanks :-)
 

Similar threads

Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 27 ·
Replies
27
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
11
Views
3K
  • · Replies 19 ·
Replies
19
Views
6K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
11
Views
8K
  • · Replies 9 ·
Replies
9
Views
3K