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Work energy theorem

  • Thread starter vissh
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  • #1
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Hello :)

Homework Statement


<Q>A wheel of moment of inertia "I" and radius "R" is free to rotate about its center as shown in figure http://s1102.photobucket.com/albums/g448/vissh/?action=view&current=pulley-1.jpg" [Broken].A string is wrapped over its rim and a block of mass m is attached to the free end of the string. The system is released from rest. Find the speed of the block as it descends through a height h.

Homework Equations


>K.e.of a body rotating abt an axis with M.O.I = I and angular velocity 'w' = (Iw2)/2
>Principle of conservation of energy [applied when work done by external forces is 0 and the internal forces are conservative]
>Conservative forces are those forces whose work done only depend on initial and final position only.

The Attempt at a Solution


>I was able to solve the problem[using a different way shown below]
>The book got a solution which i can't understand :-
It said (let at that instant) velocity of block is "v" and thus, its k.e. is (mv2)/2. Thus, angular speed of the pulley is "v/r" and thus its K.E. is [I(v/r)2]/2 .
Using principle of conservation of energy :-
..................... mgh = (mv2)/2 + [I(v/r)2]/2
>The problem is that I can't get How "tension" in string is considered to be conservative and if it is conservative, isn't a P.E. also be stored w.r.t. it .

Homework Statement





Homework Equations





The Attempt at a Solution

 
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Answers and Replies

  • #2
kuruman
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Tension is an internal force to the mass-pulley system. You can work out for yourself that it does negative work on the hanging mass and positive work on the pulley such that the sum of these two works is zero.
 
  • #3
tiny-tim
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hello vissh! :smile:
The problem is that I can't get How "tension" in string is considered to be conservative and if it is conservative, isn't a P.E. also be stored w.r.t. it .
yes, as kuruman :smile: says, the tension does no work

yes, technically a PE is stored in it, but that PE is amazingly small …

string is virtually inextensible, ie its k is extremely small, and can be ignored, in exactly the same way as we also ignore the mass of the string, the torsion (twisting) of the string, the friction on the bearings, etc

you have to accept that exam questions are deliberately worded so as to make the answer easy!!

don't read things into the question that aren't there, and do accept without question such conventions as "strings have no mass and no PE" :wink:
 
  • #4
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Thanks to you both for reply ^.^
You can work out for yourself that it does negative work on the hanging mass and positive work on the pulley such that the sum of these two works is zero.
The thing is this :-
->This was an example in my book and its was given before the topic :- "Work done by a torque"
. SO, i wasn't able to find the work done by the torque [produced by the tension] on the pulley
and thus, wasn't able to see that the work done by tension is zero in the system.
>So, we have to see if work done by tension is zero or not [in system] before applying the equation i wrote in my first post ???

:D
 
  • #5
tiny-tim
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vissh, you're doing it again! :rolleyes:

you're finding things that aren't there, and worrying about them …

you're the first person to mention torque

(yes, of course there is a torque on the pulley, but …)

kuruman :smile: only mentioned the work done by the tension … the tension is not a torque (measured in N.m), it's only a force (measured in N)! …

you should know that work done equals force times displacement …

since the string is inextensible, the displacements at opposite ends of any section of the string are the same, and since the tension forces are equal and opposite (for a "weightless" string), the work done cancels

no torque involved! :smile:

>So, we have to see if work done by tension is zero or not [in system] before applying the equation i wrote in my first post ???
no, the work done by a massless string is always zero (and for a string with mass, the work done equals the change in gravitational PE of the string, so again you can ignore it)

internal forces cancel (that's Newton's third law ), and their work done cancels … you can always ignore them :wink:
 
  • #6
kuruman
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. SO, i wasn't able to find the work done by the torque [produced by the tension] on the pulley and thus, wasn't able to see that the work done by tension is zero in the system.
>So, we have to see if work done by tension is zero or not [in system] before applying the equation i wrote in my first post ???
:D
It seems that you are using the term "system" a bit loosely. Before doing any calculations, you should be clear on what you have chosen as your "system."

1. If your system is only the pulley, then you will have to calculate the work done by the tension because it is external to the system. This calculation involves the work-energy theorem, ΔKEpulley = Wtorque
Similar considerations apply if your system is the hanging mass only.

2. If your system is the pulley and the hanging mass together, you cannot go wrong if you use the work-energy theorem and say

ΔKEpulley+hanging mass = Wgravity

If you do that, there are two points to be made:

(a) As noted earlier, the tension does equal and opposite amounts of work on the two components of this system, therefore it appears as zero on the right side and is ignored.

(b) The right side can be moved to the left side to give
ΔKEpulley+hanging mass - Wgravity = 0
But ΔPEgravity = - Wgravity by definition so you end up with the mechanical energy conservation equation
ΔKEpulley+hanging mass + ΔPEgravity = 0

This last equation is essentially the equation you wrote in your first post.
 
  • #7
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heya thanks Both of you again for replying ^.^
[Sorry for late response But my Internet went down for 2 days]:wink:
vissh, you're doing it again! :rolleyes:

you're finding things that aren't there, and worrying about them …

you're the first person to mention torque

(yes, of course there is a torque on the pulley, but …)
Hehe . Thanks for pointing out :D
And btw got it now :)

-----> Btw an extra question :- If There is a pulley as in my question and a force is acting on it as the tension T . Then, the work done by T [with pulley as the system]is equal to the work done by the torque it produces ???

Thanks both of you again :)
 
  • #8
kuruman
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heya thanks Both of you again for replying ^.^
[Sorry for late response But my Internet went down for 2 days]:wink:

Hehe . Thanks for pointing out :D
And btw got it now :)

-----> Btw an extra question :- If There is a pulley as in my question and a force is acting on it as the tension T . Then, the work done by T [with pulley as the system]is equal to the work done by the torque it produces ???

Thanks both of you again :)
I think it is better to think of it this way, there is no work done by the force of tension because the pulley is not displaced. There is only work done by the torque generated by the tension.
You know that Wtorque = τθ. Now τ = TR and if the mass descends by amount h, θ = h/R. Putting it together gives Wtorque=(TR)(h/R) = Th

The work done by the tension on the hanging mass is -Th because the displacement and the tension are at 180o with respect to each other.
 
  • #9
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I think it is better to think of it this way, there is no work done by the force of tension because the pulley is not displaced. There is only work done by the torque generated by the tension.
So, when i write the work energy theorem [taking pulley as system {or a system like this pulley (more generally)}] , I should include work done by the tension , Wtension And work done by torque , Wtorque "separately" ?
And the later part you wrote Got it :D [Amusingly, i solved like that. But i was taking Wtension on pulley = Wtorque = (Tr)(h/r) = (T)(h) ]{But as i asked above , the 2 works will be different from each other rite !!}
 
  • #10
kuruman
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So, when i write the work energy theorem [taking pulley as system {or a system like this pulley (more generally)}] , I should include work done by the tension , Wtension And work done by torque , Wtorque "separately" ?
And the later part you wrote Got it :D [Amusingly, i solved like that. But i was taking Wtension on pulley = Wtorque = (Tr)(h/r) = (T)(h) ]{But as i asked above , the 2 works will be different from each other rite !!}
You didn't quite get it. One more time,

1. When your system is only the pulley there is only work done by torque. The tension does not do work.
2. When your system is only the hanging mass, there is work done by both tension and gravity.
3. When your system is the combined pulley and hanging mass, there is only work done by gravity.

Happy holidays.
 
  • #11
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hehe my bad . got it :)

Happy holidays to you too ^.^
 
  • #12
tiny-tim
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hi kuruman! merry fishmas! :smile:
I think it is better to think of it this way, there is no work done by the force of tension because the pulley is not displaced. There is only work done by the torque generated by the tension.
You know that Wtorque = τθ. Now τ = TR and if the mass descends by amount h, θ = h/R. Putting it together gives Wtorque=(TR)(h/R) = Th
surely it's not the displacement of the pulley that matters, but of the point of application of the force?

if we consider the pulley, together with a particular bit of string, then the point of application of the tension is the end of that bit of string, which will move a distance h …

so the work done by the tension is Th …

the same as your torque result, but obtained much more directly

(i think you were thinking of friction on a rolling wheel, where there's no string to add to the pulley, and the point of application really is undisplaced, and no work is done :wink:)
 
  • #13
kuruman
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hi kuruman! merry fishmas! :smile:


surely it's not the displacement of the pulley that matters, but of the point of application of the force?

if we consider the pulley, together with a particular bit of string, then the point of application of the tension is the end of that bit of string, which will move a distance h …

so the work done by the tension is Th …

the same as your torque result, but obtained much more directly

(i think you were thinking of friction on a rolling wheel, where there's no string to add to the pulley, and the point of application really is undisplaced, and no work is done :wink:)
Correct-o, but I considered the system as being pulley sans string. Anyway, it's clear what's what.

Merry fish-mas to you too tiny-tim. Tonight's your special night - stay clear of scrooges until the morn. :smile:
 

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