A balls velocity when it reaches the top of the inside of a loop

  • #1

Homework Statement


A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.08 m/s on a horizontal section of a track as shown in the figure below. It rolls around the inside of a vertical circular loop of radius r = 45.8 cm.
given radius =0.458m
ω= 4.08

Homework Equations


KE1 + GPE1 = KE2 +GPE2
V = Rω


The Attempt at a Solution


I did the algrabra and ended up with Vf = ( (ωR)^2 -4gR)^(1/2)
i end up with a negative number.
 

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Answers and Replies

  • #2
tiny-tim
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hello jeff! :smile:

what formula did you use for KE ?

anyway, you need to show us your full calculations if we're to spot your mistake :wink:
 
  • #3
i used 0.5mv^2

KE1 + GPE1 = KE2 + GPE2
GPE1 = 0
so then i got
0.5mV1^2 = m(0.5V2^2 + gh) h = 2R (masses cancel each other out)
V1^2 - 2gh = V2^2
V2 = ((ωR)^2 -4gR)^(1/2)
V2= ((4.08(m/s)*0.458m)^2 - 4*9.8(m/(s^2))*0.458m)
 
  • #4
tiny-tim
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i used 0.5mv^2

(try using the X2 button just above the Reply box :wink:)

ah, you need to add the rotational kinetic energy :smile:
 
  • #5
but wouldn't i need to add inertia to the equation?
 
  • #6
tiny-tim
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not following you :confused:
 
  • #7
KErot = 0.5Iω^[2]
is that not right
since is do not have the mass of the ball i wouldn't not be able to solve equation
because i would not be able to cancel out inertia
 
  • #8
tiny-tim
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  • #9
im still getting the wrong answer
instead of having
V2 = ((ωR)^2 -4gR)^(1/2)
i now have V2 = (((ωR)2 -4gR)/R2)1/2
which ends up with a wrong answer
 
  • #11
haruspex
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im still getting the wrong answer
instead of having
V2 = ((ωR)^2 -4gR)^(1/2)
i now have V2 = (((ωR)2 -4gR)/R2)1/2
which ends up with a wrong answer
I don't know how you get that equation. You need to show all your working.
If the ball has radius r, what is its moment of inertia? Rolling at speed v, what is its rotational KE? What is its total KE?
 
  • #12
I figured it out
I needed my KErot to be 2/5Iw^2 and realized that
Vi = 4.08 not w =4.08
w = V/R


Sorry I typed this on my phone
 
  • #13
haruspex
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I figured it out
I needed my KErot to be 2/5Iw^2 and realized that
You mean KErot = (1/2)Iw2 = (1/2)(2/5)mr2w2 = (1/5)mv2, right? Except, that's still wrong. This is a hollow sphere.
 

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