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

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Homework Help Overview

The discussion revolves around the dynamics of a tennis ball rolling inside a vertical circular loop. The problem involves concepts from rotational motion and energy conservation, specifically relating to kinetic and potential energy as the ball moves through the loop.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of energy conservation principles, questioning the correct formulation of kinetic energy for both translational and rotational motion. There are attempts to derive the final velocity at the top of the loop, with some participants expressing confusion over their calculations and the inclusion of rotational kinetic energy.

Discussion Status

Several participants have provided insights into the equations used, with some suggesting the need to incorporate rotational kinetic energy into the calculations. There is an ongoing exploration of the moment of inertia and its impact on the overall energy equations. While some participants have made progress in their understanding, there remains a lack of consensus on the correct approach and calculations.

Contextual Notes

Participants note the absence of mass in their calculations and the implications this has on solving the equations. There is also mention of specific values for velocity and radius, which are critical to the problem but may not have been fully utilized in the discussions.

jeff.reinecke
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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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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:
 
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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
 
  • #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.
 

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