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Ball going around loop

  • #1

Homework Statement


a ball with:

mass- .06722kg
diameter- 2.53cm

rolling down a track of length 1.26m.

Find a) the minimum height the track must be for the ball to go around the loop. Loop has a diameter/height of .3m

Like this:
1m40209jicopy2.jpg




The Attempt at a Solution


I did conservation of mechanical energy with just the loop by doing KE total(bottom) = PE(top) to find the initial velocity of the ball before going up loop, giving me a tangential velocity of 1.879 m/s. I used this number to put into the kinematics equation vf=sqrrt(vi^2 + 2ad), to give me an acceleration of 1.401 m/s^2. I then used dynamics F=ma to get a force of the ball going down the slope of .09418 N, used the weight force to get a Fwt = .6594 N, and set up a triangle to find theta (call it @ here). did sin@=.09418/.6594, and solved to get a @ of 8.21 deg. I then put that angle back into the first ramp, and used trig to get a height of .18 m, or 18 cm.

If you can follow that, I have 2 questions:

1) Did i do this problem right, or am i all wrong?

2) When I try to solve for friction, I use the conservation of ME formula, to get PE - Ff = KE...however when I plug my numbers back in, I get a bigger number on the KE side then the PE side...which leads me to think I did this problem wrong in the first place? I've been working on this for about 1.5 hours now and can't seem to figure it out..past what I have done already. :( ...and yes I did add rotational KE to my totatl KE when I did the problem...
 

Answers and Replies

  • #2
23
0
OK, so at the top of the loop Fg <= Fc for it to not fall.

So if we say...

Fg= Fc
mg=mv^2/r
g=v^2/r
v=[tex]\sqrt{gr}[/tex]

Once we know the velocity it needs to reach at the top of the loop....

Ek(at top of loop)=Eg(at top of ramp)
1/2mv^2 = mgh
v^2=2gh
[tex] \sqrt{gr}^2=2gh[/tex]
gr=2gh
r=2h
h=r/2
h=0.3/2
h=15cm

.....I'm probably wrong though. It would make sense that the value needs to be greater than 0.3, simply because of conservation of energy. Good problem, interested in solution.

Thats how I would have attacked it though
 
Last edited:
  • #3
EDIT: yea, thats why i thought i was wrong in the beginning...oh well. thanks for your attempt tho. :) I appreciate it.

...anyone else?
 
  • #4
23
0
Yeah, either way it doesn't make sense. The height has to be at least 0.3m. Hmmmm I like this.

I think I have it 1 sec.
 
Last edited:
  • #5
23
0
Fg= Fc
mg=mv^2/r
g=v^2/r
v=[tex]\sqrt{gr}[/tex]

Eg(initial)= Eg(at top of ramp) + Ek
mgh1 = 2mgr + 1/2 mv^2
gh1=2gr + 1/2 v^2
gh1= 2gr +1/2 rg (subbed in earlier value of v)
h1=2r +1/2r
h1= 2.5r
h1= 2.5(0.15)
h1=37.5cm

Maybe?

The ball will have gravitational potential energy AND kinetic energy, in the first attempt, I only said it had kinetic energy. Let me know if my work is to messy and I'll latex it.

Can anyone confirm this solution???
 
Last edited:
  • #6
OHHH...it would make sense that the ball would have BOTH KE and PE at the top of the loop! I never took that into consideration! I think you are right! :D
 

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