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

AI Thread Summary
The discussion centers around calculating the velocity of a tennis ball at the top of a vertical loop, starting with an initial speed of 4.08 m/s. Participants emphasize the importance of incorporating both translational and rotational kinetic energy in the calculations, specifically noting the moment of inertia for a hollow sphere. One user struggles with the algebra and ends up with incorrect equations, prompting others to encourage showing full calculations for clarity. The correct approach involves recognizing that the rotational kinetic energy should be expressed as (2/5)Iω², leading to a realization that the initial velocity should be treated as linear speed rather than angular speed. Ultimately, the conversation highlights the need for careful application of physics principles in solving the problem.
jeff.reinecke
Messages
11
Reaction score
0

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.
 

Attachments

  • photo.jpg
    photo.jpg
    46.5 KB · Views: 679
Physics news on Phys.org
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:
 
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)
 
jeff.reinecke said:
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:
 
but wouldn't i need to add inertia to the equation?
 
not following you :confused:
 
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
 
jeff.reinecke said:
KErot = 0.5Iω^[2]

yes :smile:

call the mass "m", and find the moment of inertia as a multiple of m

(m will cancel out in the end)

and now I'm off to bed :zzz:
 
  • Like
Likes 1 person
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
 
  • #10
please help anybody
 
  • #11
jeff.reinecke said:
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
jeff.reinecke said:
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.
 
Back
Top