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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Dec 2007 22:43:25 IST
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How can energy be conserved in this case: a thin rod is hinged at its end and is vertical initially .it is pushed slightly and allowed to rotate about the horizontal axis through its end there is a hinge force acting on it which is NOT along the rod(for instance when the rod makes an angle of 90 degress with the vertical,its angular acceleration is less than g therefore some upward force must be acting on it that isnt along the rod. Therefore it will do some work.) So how can we conserve energy?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Dec 2007 23:25:00 IST
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the hinge force wont do any work as it is suspended from a stationary pt.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Dec 2007 02:42:37 IST
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Yes net work by that force is zero as displacement of point of application of force is zero
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Dec 2007 08:26:53 IST
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um..the hinge applies a force in both horizontal and vertical directions on the COM.Also the net force isnt along the rod...so y shud energy b conserved?
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now here there should be a clear distinction between whether u r asking for work done by torque or work done by force. if torque then about which point.if u choose the point as the hinged point then clearly there is no work done by torque due to hinge because the torque itself is 0. coming to work done by force then whether u want the work done on that point or the centre of mass . if that point then work done is 0 .because that point itself is not moving
K=0 so the work done by net force should be 0. if u consider centre of mass then this force has two components one tangential and the other radial.The radial one does no work for obvios reasons while the tangential one does some work
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Dec 2007 10:28:53 IST
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if we ahve to calculate the velocity of the COM when the rod falls down,we use Conservation of energy.ow mg provides the torque to the rod.So is using COE correct?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Dec 2007 11:32:07 IST
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yes .but i would like to call that the work-energy principle for torque and that follows from the definition of rotational kinetic enrgy.Work done by the trque due to mg about the hinge is equal to total change in rotational k.e of the rod about hinge since no other external torque acts on the rod about the hinge
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Dec 2007 13:47:02 IST
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thnx
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