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Mechanics

New kid on the Block

 Joined: 22 Aug 2012 Post: 2
22 Aug 2012 22:41:33 IST
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A rod of length L is pivoted at one end and is rotated with uniform angular velocity in a horizontal plane.Let T1 and T2 be the tensions at the points L/4 and 3L/4 away from from pivoted ends.find the relation between two tensions....[DOUBT:WHAT WILL PROVIDE REQUIRED CENTRIPETAL FORCE?]

Blazing goIITian

Joined: 4 Jan 2007
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24 Aug 2012 00:43:26 IST
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dm = (M/L)dx

T - (dT+T) = (dm) W^2 x

-

Now put value of  x = L /4 and 3L/4 and get the result

Blazing goIITian

Joined: 4 Jan 2007
Posts: 489
25 Aug 2012 20:53:22 IST
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Tell me whether u got the right answer

Forum Expert
Joined: 29 Dec 2006
Posts: 5535
28 Sep 2012 00:25:14 IST
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Tension in the rod will provide required centripetal force. Due to mass of rod this tension will have different value at different location.

New kid on the Block

Joined: 1 Oct 2012
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1 Oct 2012 13:22:43 IST
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tension in the rod will decrease as the distance from center increase??

New kid on the Block

Joined: 1 Oct 2012
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1 Oct 2012 13:34:49 IST
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i think it will increase as the centripetal force will increase as the distance increase...please comment on this

Cool goIITian

Joined: 16 Sep 2011
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4 Oct 2012 20:51:28 IST
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n = a+b so s = x + y so integral z = x therefore x = L/4 and 3L/4.........this method is direct

Cool goIITian

Joined: 16 Sep 2011
Posts: 65
4 Oct 2012 20:52:18 IST
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therefore dm = (M/L)dxT - (dT+T) = (dm) W^2 x-Now put value of x = L /4 and 3L/4 and get the result

New kid on the Block

Joined: 9 Oct 2012
Posts: 1
9 Oct 2012 07:19:05 IST
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dont post silly questions like dis ...plz...!!!!!!!!

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