Rotation and Elongation of a spring 1

A very long thin tube is hinged at a point such that its free to rotate in the horizontal plane. Inside the tube there is spring fixed at its end. the spring is connected to a mass m m which precisely fits into the tube.

At the instant t = 0 t=0 when the spring is in its natural length of R R , the mass is given velocity V V in a direction perpendicular to the length of the tube.

If the maximum elongation produced in the spring is X X then find the value of ( 100 X 2 + 99 ) ( X + 2 ) (100X^2 + 99)(X+2) .

In this case, m = 2 kg , V = 1 m/s , k = 200 N/m , R = 1 m m = 2 \text{ kg}, V = 1 \text{ m/s}, k = 200 \text{ N/m}, R = 1 \text{ m} .

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The answer is 200.

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1 solution

Yash Sharma
Apr 4, 2015

nice solution...

manish bhargao - 6 years, 2 months ago

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thanks............this is part of my set spring physics..try my other problems too

Yash Sharma - 6 years, 2 months ago

Did the same!!!! Great question

A Former Brilliant Member - 4 years, 5 months ago

U did a wromg calculation, during the transition of pages, u wrote 100x in place of 100x^2 , no wonder i was not getting the right ans.....

Raunak Agrawal - 4 years, 3 months ago

yes i do agree with @Raunak Agrawal

M@dhur T. - 4 years, 3 months ago

Hey, I have a doubt. Apart from angular momentum conservation, can we use k*x = mv^2/r instead of energy conservation? If not, then why?? Thanks!

Brilliant Member - 3 years, 3 months ago

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