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Titun's Avatar
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30 Dec 2006 18:42:25 IST
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ROTATIONAL MECHANICS
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A circular wheel has a mass of m. It has 12 spokes each of which has a mass of m/4. The wheel is rotating with an angular speed of w. What is the amount of work done in stopping the wheel?


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ace's Avatar

Cool goIITian

Joined: 11 Dec 2006
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30 Dec 2006 21:01:45 IST
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is the question complete???????????? something seems amiss

Cool goIITian

Joined: 10 Dec 2006
Posts: 47
31 Dec 2006 13:02:11 IST
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The moment of inertia about the axis
= MI due to ring (rim) + MI due to spokes
= m * r^2 + 12 * 1/3 * m/4 * r^2
=2 * m * r^2

The work done to stop will be equal to the (initial rot.KE - final KE )
= 1/2 * (2 * m * r^2 ) * w^2
= m * r^2 * w ^2

New kid on the Block

Joined: 31 Dec 2006
Posts: 4
31 Dec 2006 22:11:44 IST
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the question does not specify the sense of rotation.........whether the wheel is just rotating about the central axis or is it rotating without slipping on a ground?????


in both the cases work done in stopping is = loss in total kE


Moderator
Joined: 29 Dec 2006
Posts: 1037
1 Jan 2007 15:36:54 IST
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Dear
  The question is incomplete we need the radius of the wheel
  Given the radius
  All you have to do is get the MOI of  the wheel ie MOI of the spokes( M/4r^2/3 + MOI of the wheel as a ring ( Mr^2 )

  Net work that need to be done can be easily found from Work energy theorom ( angular )
 i.e Change in Rotational KE = Work done ( Becareful about the torque check if its constant or variable )

in your case I have assumed that torque is constant and change in rotational KE  is simply initial KE  since finally the wheel is coming to rest
Tarasha Dewan's Avatar

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Joined: 21 Dec 2006
Posts: 38
1 Jan 2007 15:38:32 IST
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hi the work done wil be equal to the change in rotational kinetic energy, that is
 
work done = final rotatinal ke - initial rotational ke.
 
since final ke is zero, we get the work done as the negative of the initial rotational ke .
 
that is work done = 0 - 1/2 I W^2
 
WHERE I is the moment of inertia of the wheel =
 mr^2  (momentof inertia of the wheel)      + 12 * 1/3 *mr^2/4      (moment of inertia of the 12 spokes which are assumed to be uniform rods of lenth r and mass m/4)
 
 
put in the values in formula of ke and solve for the answer
 
cheers!!
 
tarasha

Moderator
Joined: 29 Dec 2006
Posts: 1037
1 Jan 2007 15:39:37 IST
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Dear
  The question is incomplete we need the radius of the wheel
  Given the radius
  All you have to do is get the MOI of  the wheel ie MOI of the spokes( M/4r^2/3 + MOI of the wheel as a ring ( Mr^2 )

  Net work that need to be done can be easily found from Work energy theorom ( angular )
 i.e Change in Rotational KE = Work done ( Becareful about the torque check if its constant or variable )

in your case I have assumed that torque is constant and change in rotational KE  is simply initial KE  since finally the wheel is coming to rest



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