Elongation in rod

Classical Mechanics Level 4

An elastic rod of costant k k and length l l is hung at a point P, rod have a massless lock at the other end. A smooth ring of mass m m falls from P (It is the same point from where the rod was hung), find the max elongation in the rod.

Details and Asumptions :

  • Gravity acts vertically downwards.

  • Position of ring in picture is not of initial position.

  • Length l is taken from point P to lock.

  • l = 30 , m = 20 , k = 10 , g = 10 l=30, m=20, k=10, g=10

  • All the values above are in SI units


The answer is 60.

Kushal Patankar by Kushal Patankar , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Satvik Pandey
Mar 21, 2015

We can easily use conservation of energy here.

From the figure

Initial energy of the system= m g l mgl

Final energy of the system= k x 2 2 − m g x \frac{kx^{2}}{2}-mgx

By conservation of energy

m g l = k x 2 2 − m g x mgl=\frac{kx^{2}}{2}-mgx

On putting values we get

x 2 − 40 x − 1200 = 0 x^{2}-40x-1200=0 So x = 60 x=60

8 Helpful 0 Interesting 0 Brilliant 0 Confused

The elongation seems a bit too high dont you think, no rod can tolerate that and still follow hooks law

Mvs Saketh - 6 years, 2 months ago

Log in to reply

Yes but I think problem maker wants us to solve this by using concept of Hooks law.

satvik pandey - 6 years, 2 months ago

Indeed, yes. But our task is to solve the problem anyway.

Rishav Koirala - 6 years, 2 months ago

I think, because the elongation is too hight, that we should use k = E S l k=\frac{ES}{l} because now the change of l l isn't that small to say k = c o n s t k=const .

Вук Радовић - 6 years, 2 months ago
  • OVER-RATED QUES
  • if x is the elongation, the ring falls through (l + x). Thus the loss in P.E.=mg(l+x). The gain in spring energy . Equate the two.
0 Helpful 0 Interesting 0 Brilliant 0 Confused

If x is the elongation, the ring falls through (l + x). Thus the loss in P.E.=mg(l+x). The gain in spring energy = 1 2 ∗ k ∗ x 2 \dfrac{1}{2}*k*x^2 . Equating the two, and solving the quadratic for +tive value, x=60. I agree that the values given are not possible. Because of this some might have avoided the problem thinking they have not understood the problem well.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...