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22 Aug 2012 22:41:33 IST

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CIRCULAR MOTION

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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?]

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yahiyafirdous

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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

yahiyafirdous

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25 Aug 2012 20:53:22 IST

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Tell me whether u got the right answer

Krishna Gopal Singh

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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.

Anubhav

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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??

Anubhav

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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

Parth Chhaparwal

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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

Parth Chhaparwal

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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

saiharsha

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9 Oct 2012 07:19:05 IST

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dont post silly questions like dis ...plz...!!!!!!!!

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