A rod of length L is pivoted at one end and is rotated witha uniform angular velocity in a horizontal plane.Let T1 and T2 be the tnesion at the points L/4 and 3L/4 away from the pivoted ends A.T2 > T1 B.T1>T2 Ans- B

Dear manoj let mass per unit length of rod is m and take a strip of lenghth dx at a distance x from the pivoted end tension in rod is due to centrifugal force in part of rod that is away from the point of interest so for point at a distance L/4 T1 = limit L/4 to L ∫ (mdx)w2 x =15/32 mw2L2 For a point at 3L/4 T2 = limit 3L/4 to L ∫ (mdx)w2 x =7/32 mw2L2 So T1>T2 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Askiitians Experts Badiuddin

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A rod of length L is pivoted at one end and is rotated witha uniform angular velocity in a horizontal plane.Let T₁ and T₂be the tnesion at the points L/4 and 3L/4 away from the pivoted ends

A.T₂ > T₁ B.T₁>T₂

Ans- B

manoj jangra , 16 Years ago

Grade 12

2 Answers

Badiuddin askIITians.ismu Expert

Dear manoj

let mass per unit length of rod is m

and take a strip of lenghth dx at a distance x from the pivoted end

tension in rod is due to centrifugal force in part of rod that is away from the point of interest

so for point at a distance L/4

T₁ = limit L/4 to L ∫ (mdx)w² x

=15/32 mw²L²

For a point at 3L/4

T₂ = limit 3L/4 to L ∫ (mdx)w² x

=7/32 mw²L²

So T₁>T₂

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Regards,
Askiitians Experts
Badiuddin

Last Activity: 16 Years ago

Rishi Sharma

Dear Student,
Please find below the solution to your problem.

let mass per unit length of rod is m
and take a strip of lenghth dx at a distance x from the pivoted end
tension in rod is due to centrifugal force in part of rod that is away from the point of interest
so for point at a distance L/4
T1 = limit L/4 to L ∫ (mdx)w2 x
=15/32 mw2L2
For a point at 3L/4
T2 = limit 3L/4 to L ∫ (mdx)w2 x
=7/32 mw2L2
So T1>T2

Thanks and Regards

Last Activity: 5 Years ago