Moment of inertia and work problem

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Homework Help Overview

The problem involves a cylindrical spool with a nylon cord being pulled, requiring the calculation of work done on the spool as it accelerates to a specific angular speed. The subject area includes concepts of rotational dynamics, moment of inertia, and energy conservation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the moment of inertia of the spool, questioning whether it is solid or hollow. There are attempts to calculate tension and angular acceleration, with some suggesting the use of angular kinetic energy formulas and conservation of energy as alternative approaches.

Discussion Status

The discussion is ongoing, with participants offering different perspectives on the calculations and methods. Some guidance has been provided regarding the use of energy conservation and the correct application of moment of inertia formulas, but no consensus has been reached on the correct approach.

Contextual Notes

There is uncertainty regarding the type of spool (solid vs. hollow) and the implications this has on the moment of inertia. Participants express confusion over the formulas involved, indicating a need for clarification on the concepts being applied.

coolbyte
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Homework Statement



A 4.00-m length of light nylon cord is wound around a uniform cylindrical spool of radius 0.500 m and mass 1.00 kg. The spool is mounted on a frictionless axle and is initially at rest. The cord is pulled from the spool with a constant acceleration of magnitude 2.82 m/s2.

How much work has been done on the spool when it reaches an angular speed of 7.95 rad/s?


Homework Equations



Angular acceleration = a/r
Moment of inertia of the cylinder is mr^2

The Attempt at a Solution



I solved for tension first: T*0.5 = 1*(0.5)^2*(2.82 / 0.5)=> So T=2.82 N

Then using kinematic equations I solved for theta:

Angular acceleration = a/r = 2.82/0.5 = 5.64

so

7.95^2 = 2 * 5.64 * theta ==> theta = 5.6

so work = 2.82 * 5.6 * 0.5 = 7.896 joules
However webassign says I'm wrong, any help would be greatly appreciated as I spent too much time into this, thanks.
 
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coolbyte said:
Moment of inertia of the cylinder is mr^2
That would be for a hollow cylinder. I would assume the spool is solid.
I solved for tension first: T*0.5 = 1*(0.5)^2*(2.82 / 0.5)=> So T=2.82 N
That's a long way round. Just use the angular KE formula, Iω2/2.
 
It looks like you calculated the polar moment of inertia incorrectly. Recheck your formula. Also, even though your approach should deliver the correct answer, it would be much easier to solve this problem by applying conservation of energy. That way, you wouldn't even have to calculate the tension.
 
haruspex said:
That would be for a hollow cylinder. I would assume the spool is solid.

That's a long way round. Just use the angular KE formula, Iω2/2.

Thanks mate, you're right, those formulas always confuse me.
 
coolbyte said:
Thanks mate, you're right, those formulas always confuse me.
mr2 applies for a point mass at distance r, so that must also be the formula for a hoop about a line orthogonal to the plane of the hoop, and for a hollow thin shell cylinder about its axis. That's because in each of those cases every part part of the body rotates at radius r from the axis. For a solid disc or cylinder, or for a sphere (solid or hollow), or for a hoop rotating about a diameter, it must be less, since no part rotates at distance greater than r but some parts rotate at a shorter distance.
 

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