TBD physics pendulum a

[capn jazz]

---

Members don't see this ad.

Chapter 5, passage 3, q17:

A pendulum is pulled a small angle theta and allowed to fall. What can be concluded about the angular displacement theta' after half a cycle has elapsed?also note the string is length L and there is a nail in the wall a distance of (3/4)L below the point of pendulum attachment.

A) theta'=theta
B) theta'> theta
C) theta'< theta

The answer is b and I can't understand why...

 

[loveoforganic]

Ok, I think I understand this question. Imagine your pendulum as a ball attached to a string. The pendulum, from the top of its arc to the center point of its swing, moves through an angle theta. At this point, a nail collides with the pendulum's string 3/4 of the way down the string. This causes you to have a new pendulum with L = 1/4 of the length of the original pendulum. Your restoring force is given by F = -mg(x/L). As L decreases, x (the displacement) must increase. The only way for x to increase is by moving through a larger angle theta'.

If that isn't right, someone please correct me

 

[capn jazz]

That makes perfect sense, thanks!

 

[redlight]

Ok, I think I understand this question. Imagine your pendulum as a ball attached to a string. The pendulum, from the top of its arc to the center point of its swing, moves through an angle theta. At this point, a nail collides with the pendulum's string 3/4 of the way down the string. This causes you to have a new pendulum with L = 1/4 of the length of the original pendulum. Your restoring force is given by F = -mg(x/L). As L decreases, x (the displacement) must increase. The only way for x to increase is by moving through a larger angle theta'.

If that isn't right, someone please correct me

im pretty sure thats right. i look at it from an energy standpoint.

the effective length of the pendulum changes once the rope swings and hits the nail (time = 1/4 of a cycle). only 1/4 of the original length of the rope is now swinging.

the energy in the system is still conserved and remains the same. the pendulum swings to the same height that you started with at t0, but since the rope is shorter, the angular displacement theta is larger.
vb = velocity at the bottom, as the pendulum rope hits the nail = sqr(2gH)
initial H in terms of theta = (L - Lcos(theta))

H at t1/2 = (L`- L`cos(theta prime)), where L` = L/4

for both H's to be the same, theta prime has to be larger.

 

[redlight]

im pretty sure thats right. i look at it from an energy standpoint.

the effective length of the pendulum changes once the rope swings and hits the nail (time = 1/4 of a cycle). only 1/4 of the original length of the rope is now swinging.

the energy in the system is still conserved and remains the same. the pendulum swings to the same height that you started with at t0, but since the rope is shorter, the angular displacement theta is larger.
vb = velocity at the bottom, as the pendulum rope hits the nail = sqr(2gH)
initial H in terms of theta = (L - Lcos(theta))

H at t1/2 = (L`- L`cos(theta prime)), where L` = L/4

for both H's to be the same, theta prime has to be larger.

i attached a pic but it didnt show up. let me try again.

 

[loveoforganic]

I like the energy reasoning more. Energy reasoning in general tends to make more sense.