Elasticity
Consider 2 similar Rods of mass M and length L .Their cross-sectional area is A and Young's Modulus is Y .They are in the following conditions.
R o d − 1 : It is hinged at one end and is rotating in the horizontal plane with constant angular velocity ω .
Its Elastic Potential Energy is U 1 = a A Y M 2 ω 4 L 3
R o d − 2 : It is hanging freely (through its one end) from ceiling
Its Elastic Potential Energy is U 2 = b A Y M 2 g 2 L
where a & b are positive integers
Enter your answer as b a
This is a part of my set Aniket's Mechanics Challenges .
The answer is 2.5.
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2 solutions
a = 15 and b = 6
First case:
Let the axis to be at left end of rod
Take an element dx at a distance x from the axis. Tension at this part will provide centripetal force to the part right to it. T = L M ( L − x ) ω 2 ( x + 2 L − x )
Now calculate U
d U = 2 A Y T 2 d x
Can you tell how a=15 as for some reasons i am getting a =60
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First find the tension in an element dx at a distance x from axis which is equal to the other part's centre of mass's centripetal force. T= M ( L 2 − x 2 ) w 2 /2LThen put it in the equation U = 0.5 T 2 dx/AY ....now did you get ?
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I did my taking an elemental mass at a distance x from hinged end now U = 2 1 × ( strain ) 2 × Y volume .
Force on elemental mass is the centripetal force for the mass to the left to it.
d U = 2 A Y L 2 M 2 w 4 × ( L − x ) 2 x 2 d x
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@Neelesh Vij – Elaborate it ...You have done something wrong but I am not able to catch it
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@Aniket Sanghi – Elaborate what part? finding d U ?
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@Neelesh Vij – Your tension first , is it coming same as mine........then yes ...finding dU part
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@Aniket Sanghi – There you go-
As U = 2 1 × ( strain ) 2 × Y volume .
⇒ d U = 2 Y 1 × ( L A M x w 2 × ( L − x ) ) 2 × A d x
⇒ d U = 2 Y A M 2 w 4 × ( L 2 x 4 . d x + x 2 . d x − L 2 x 3 . d x )
On integrating and solving:
U = 6 0 Y A M 2 w 4 L 3
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@Neelesh Vij – Sorry to be late........... I can't understand how can you calculate strain first.......I learnt that you can calculate tension first
@Neelesh Vij – I am uploading my solution
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@Aniket Sanghi – That's the problem, i can't upload the image relevant to it so it might be a bit confusing.
I took a small element of length dx, at a length x from the end rotating, so force on particle is the force required by the part on the right to it of length (L-x) to do circular motion (centripetal force) so that is how i calculated strain
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@Neelesh Vij – Check up my solution
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@Aniket Sanghi – Why did you take the part of rod on right of dx to rotate from its C.O.M?
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@Neelesh Vij – For left part ...There are two forces acting tension and force due to axis
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@Aniket Sanghi – Can we first calculate the total extension in ROD 1 and the give answer as 1/2 kx^2
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@Sathyam Tripathi – I don't think that would work you may check it by cross verifying
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@Aniket Sanghi – It didn't work. Can you tell why
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@Sathyam Tripathi – Which form of formula of U did you use
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@Aniket Sanghi – 1/2 kx^2 where k =AY/L
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@Sathyam Tripathi – No you can't integrate first and find x first as energy is not uniform throughout, check my solution
@Aniket Sanghi – @Aniket Sanghi You there bro?
Ok i got my mistake, thanks !
solution by @Aniket Sanghi is perfectly fine, but i will just write how he got it in a bit detail , Take an element dx at a distance x from the axis. Tension at this part will provide centripetal force to the part right to it. since the mass density is constant , the mass to the right of it would be ρ (l-x) and then when we apply the centrifugal for on the c.o.m of the left part , we get its co-ordinates as given in aniket's solution and then using the simple equation of young's modulus , for the first case we get answer as (8/15) 1/8 = 1/15 some other values and the second one is easier write similarly , you get the other as 1/2 1/3= 1/6 , when you divide , them you get 5/2 = 2.5 ! :)