A pendulum with a ball and a string falls to wrap around another peg

[emmy]

Homework Statement

In the Figure 8-36, the string is L = 220 cm long, has a ball attached to one end, and is fixed at its other end. A fixed peg is at point P. Released from rest, the ball swings down until the string catches on the peg; then the ball swings up, around the peg. If the ball is to swing completely around the peg, what value must distance d exceed? (Hint: The ball must still be moving at the top of its swing. Do you see why?)

http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c08/fig08_38.gif

2. The attempt at a solution

I thought I had this one, and it ended up being wrong...
L=2.2m
Before pendulum is released:
Ug₀=mgL
Ug₀=KE_f=.5mv²=mg(2r)

where KE_f is the kinetic energy at the lowest point, and r is the radius of the circle the pendulum makes around the second peg.

mgL=mg2r
L=2r
L/2=r
2.2m/2=1.1m=r
so d (being the remaining cord length) is
L-r=d
d=2.2m-1.1m=1.1m

Where did I go wrong

 

[diazona]

First: what is your initial situation? What is your final situation?

Then: why did you write this equation?

Ug₀=KE_f

 

[emmy]

First: what is your initial situation? What is your final situation?

Then: why did you write this equation?

Since there are only conservative forces acting on the system,
0=ΔKE+ΔU
0=KE_f-KE₀+(U_f-U₀)

or

U₀-U_f = KE_f-KE₀

The potential energy at the beginning, U₀ = U_g = mgL
and U_f=0 at the bottom of the swing
The Kinetic energy at the beginning, KE₀=0 since the pendulum is not moving
The Kinetic energy at the bottom of the swing is moving at its maximum speed, and KE_f=1/2mv²

so

mgL - 0 = 1/2mv² - 0
mgL = 1/2mv²
gL=1/2v²
2gL=v²
solve and get velocity of the string at the bottom of the swing

Next, the pendulum is going to begin to swing around the smaller peg...
In order to just make it over the peg, the KE at the top of the swing must be > 0
Again:
0=ΔKE+ΔU
0=KE_f-KE₀+(U_f-U₀)

KE_f>0
KE₀= 1/2mv² (<--- this velocity is what I found above)
U_f=mg2r where r is radius
U₀= 0

U₀-U_f = KE_f-KE₀
0 - mg2r = -.5mv²
2gr=.5v²
r = .25v²

but that also gives the same answer...

 

[diazona]

True, but this time you were able to properly justify it ;) So how do you know this is the wrong answer?

 

[emmy]

True, but this time you were able to properly justify it ;) So how do you know this is the wrong answer?

Because the online homework program said so :Y

 

[diazona]

Hm, well I don't immediately see anything we're missing. I'll see if I can get someone else to take a look at it.

 

[emmy]

Hm, well I don't immediately see anything we're missing. I'll see if I can get someone else to take a look at it.

Hmm, maybe it's software malfunction?

*crosses fingers*

 

[vela]

The ball has to be moving at a minimal speed at the top, otherwise it won't make a complete loop. That minimal speed isn't 0.

At the slowest speed, the string just goes slack. You can use that criterion to figure out what v_min is.