# 2. Advance Illustration | Work Energy Power | Speed of a Falling Rope Supported on a Nail in this illustration we are, going to calculate
the speed of a falling rope supported by a nail. here the figure shows a uniform smooth
and flexible rope of mass m and length 2 l. which is symmetrically supported on a nail
ay, here if due to a slight jerk rope starts falling. we are required to find its speed
when it leaves off the nail. here in solution, we can see. if jerk is provided on the right
part and say it starts falling. and if this is nail ay and 1 rope will be, moving off
the nail. the situation would be this its total length 2 l, is on 1 side of the nail.
and here if we just have look on it center of mass this is, the center of mass, of this
rope. which is at a distance, l bellow the nail. earlier if we have a look the center
of mass of rope would be at the midpoint as rope is uniform. and the center of mass is
at a depth l by 2, from the nail ay, so here we can see, in, the process of motion. of
rope. the displacement of center of mass. of rope is. this h, by which the center of
mass is fallen down can be written as l by 2. so we can directly write using work energy
theorem. as g is uniform we can consider, the work done by gravity on the rope. in the
process of motion we’ll be m g h. so initial kinetic energy of the rope was zero. plus
work done on it is, m g l by 2. must be equal to the final kinetic energy which is half
m v square. where v is the speed attain by the rope as., from the initial moment it was
starting from rest. so, m gets cancelled out this 2 gets cancelled out and the value of
sped we are getting is root g l. that will be the final answer for this problem.

1. Vedant Patil says:

sir advance illustrations are very useful

2. Shubham Dhull says:

sir the height fallen by c.o.m. should be l as the total length is 2l .

3. Syed Sumaid says:

Sir,what about the work done by force exerted by nail?

4. Harsh says:

sir you handwriting is too good. no word for you knowledge and efforts, you gave me new hope to study physics.

5. Devansh G says:

6. Ankit Jaiswal says:

sir why we consider centre of mass of two parts of "l" lenght as "l/2" why we cannot consider that initially centre of mass of complete rope of lenght "2l" was above the nail and in this case displacement of centre of mass will be "l"? what is mistake in this?

7. Nischay Shishodia says:

Dear sir i am not able to understand how the displacement of centre of mass is l/2.
**pls explain where is initial centre of mass.
**is initial centre of mass between the ropes at l/2 and how
confused pls explain sir ji??

8. Nischay Shishodia says:

where was initial centre of mass?

9. yuvraj singh says:

Why you have not taken h=2l because mg force is face by whole rope

10. Kushagra Up says:

why did we take com in this case.

11. Kushagra Up says:

if we normall y took shift in pe

12. Kushagra Up says:

if we normally took shift in potential energy then equation is mgl=1/2mv^2

13. M.A. choudhury says:

sir, I have a simple doubt, when the rope leaves of nail and it will become vertical, so why don't we take the final P.E. ? sir plz answer

14. Raj Kumar Padha says:

15. mohit nayak says:

sir,why we are here taking the case of com.

16. ashutosh KUMAR says:

sir h=l hona chahiye na

17. Repudi Rajasekhar says:

Sir…. see so many people got doubt about lenght so explain step by step and very clearly… thank u

18. Suryansh Kumar says:

How did you Calculate displacement h/2.

19. ejaz ahmad says:

Sir is this possible yo calculate work by gravity by taking elemental part dx and then intergrate it from o to l/2 because as mass falling gravity force changes (variable force)

20. Sukhchen Singh says:

Sir please tell the limits of integration

21. Raj Das says:

Sir can we solve this by using potential energy concept (dw(conservative)=-du ) …by finding the potential energy and then solve this accordingly .

22. ABHINAV says:

Sir When the rope leaves the nail then the final potential energy=mgl why didn't you consider that pl. answer……!!!!

23. Harsh Kumar Singh says:

Initial com of the system should be at A na sir ??? 😛😛