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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Mar 2008 21:57:15 IST
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a thin uniform rod of mass m and length l rotates with constant angular velocity omega about the vertical axis passing through the rod's suspension point O. In doing so the rod describes a conical surface with half aperture angle 'theta'. what is the component of reaction force at O in x direction? and magnitude of angular momentum of rod about axis of rotation ? ans; 1. ml  2 / 2 2. ml 2 / 12
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24 Mar 2008 01:30:00 IST
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If the radius of the circle at the bottom u have drawn is also l, then d ans is rt.
if so,
Centre of mass of rod rotates in a circle of radius l/2.
Force for this can be provided only by the horizontal component of hinge force.
m 2l/2 is the centripetal force, n this is equal to horizontal component of hinge force.
I dont think magnitude of angular momentum is m l2/12.
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24 Mar 2008 02:17:10 IST
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elastiboysai if it rotates with a radius of l then it won't become a conical pendulum as far as ans are concerned i think they are wrong 2) calculating MOI alon O ml^2/12 ml^2/4 =ml^2/ 3 I = ml^2/3 |
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24 Mar 2008 02:20:32 IST
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well balganesh
I is not ml^2 / 3
if u notice the axis is passing through the end but is not perpendicular to it
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Impossible To be Impossible is Impossible |
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24 Mar 2008 10:50:42 IST
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yes balganesh..the answer by u isnt correct...
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24 Mar 2008 11:44:38 IST
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let da force thru da rod b T ............
da radius of the circle described by da rod at da bottom wud b = l sin theta
T sin = m2 l sin
T cos = mg
T = m 2 l
dis wud b da reaction force exerted at o
its component in x direction m2 l sin.
2)
moi = ml2 /3 sin
angular momentum = ml2 /3 sin
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24 Mar 2008 20:28:55 IST
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@ anki016-Are you sure components of the moment of inertia can be taken? Can u plz validate ur point through some theorem probably?? |
______________________________________________
Siddhant Shah
The trouble with doing something right the first time is that nobody appreciates how difficult it is.
You can't control the wind, but you can adjust your sails.
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24 Mar 2008 20:30:50 IST
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sid anki has merely said the radius of circle described=lsin
its not I
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24 Mar 2008 20:35:20 IST
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No elasti, not the radius, I'm talking about the second part
MoI = ml2/3sin@
that one......
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______________________________________________
Siddhant Shah
The trouble with doing something right the first time is that nobody appreciates how difficult it is.
You can't control the wind, but you can adjust your sails.
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