# Rod in Circular motion

Hello >:D

## Homework Statement

<Q>A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let T1 and T2 be the tensions at the points L/4 and 3L/4 away from the pivoted ends respectively.
<a> T1>T2 <b> T1<T2 <c> T1=T2 <d> The relation b/w T1 and T2 depends on whether the rod rotated anticlockwise or clockwise

## Homework Equations

For a pt mass "m" moving in a uniform circle. Fcentripetal = mv2/r

## The Attempt at a Solution

Hmm shouldn't the tension across the whole rod should be same and thus T1 =T2 .
Need little guidance ^.^

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ehild
Homework Helper
Look after "tension". What does it mean?

ehild

gneill
Mentor
Hmm shouldn't the tension across the whole rod should be same and thus T1 =T2 .
Need little guidance ^.^
Suppose you hang a rope from a hook. This rope is not a massless, ideal rope, but a real one with mass per unit length. Do you expect the tension to be the same near the top (where all the weight of the rope lies below and is being held up) as at the bottom where there's nothing below to support?

Thanks to you both for replying :)
>Got this definition in my book :-
Consider a rope in hanging from ceiling. Consider a cross-section of the string at any pt A in the string.This cross-section divides the string in 2 parts-lower and upper.The lower part will exert a downward force on upper one and the upper part will exert an upward force on the lower one.According to newton's third law,these two forces will have equal magnitude.The common magnitude of the forces exerted by the 2 parts of the string on each other is called "Tension" in the string at A.
>On reading it again,The tension will not be same in a string with mass and either in a rod hanging vertically. But still not get how will be it different in a horizontal clamped rod !!!

gneill
Mentor
If the rod is rotating then there will be centripetal acceleration which will vary with distance from its pivot point. This is analogous to hanging in an acceleration field (like gravity). Of course, for small displacements near the surface of the Earth, the acceleration due to gravity is nearly constant (g). The rod is going to experience an acceleration field that goes from zero at its pivot to some maximum value at its end which depends upon the rotational velocity.

thanks again :D
Hmm, Let the mass of any pt particle of the rod is "m".And the rod is rotating with "w".
So, at a pt L/4 distance away from the pivoted end, T1 = (mw2L)/4
And at a pt 3L/4 distance away from the pivoted end, T2 = 3(mw2L)/4
So i am getting T1 < T2 .
But the answer is T1 > T2 .
Where am i getting wrong ?? O.O

ehild
Homework Helper
Consider a small piece of the rod at distance l from the pivot and of length Δl. The mass of the piece is Δm, proportional to Δl. The tension of one end is T(l), and it is T(l+Δl) at l+Δl, at the other end. T(l) points toward the pivot and T(l+Δl) points away from it. What is the net force acting on this piece of the rod?
The piece moves around a circle, so the net force is equal to the centripetal force. Write the equation and see if dT/dl is positive or negative.

ehild

gneill
Mentor
thanks again :D
Hmm, Let the mass of any pt particle of the rod is "m".And the rod is rotating with "w".
So, at a pt L/4 distance away from the pivoted end, T1 = (mw2L)/4
And at a pt 3L/4 distance away from the pivoted end, T2 = 3(mw2L)/4
So i am getting T1 < T2 .
But the answer is T1 > T2 .
Where am i getting wrong ?? O.O
You need to sum up the contributions of all the little masses (dm) that are further from the point you are interested in. It's equivalent to adding up the weight of the rope hanging below a given point. A little calculus is in order.

Thanks Both of you :D Got the answer (^.^)