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

In summary: What is the moment of inertia of a hollow sphere?In summary, a tennis ball was set rolling without slipping at a speed of 4.08 m/s on a horizontal track. It then rolled around a vertical circular loop with a radius of 0.458 m. The equation used for solving this problem was KE1 + GPE1 = KE2 + GPE2, and the final formula for finding the velocity at the bottom of the loop was V2 = ((ωR)^2 -4gR)^(1/2). After some calculations, it was determined that the moment of inertia for a hollow sphere is 2/5mr^2 and the final formula for finding the total kinetic energy was KE = (
  • #1
jeff.reinecke
11
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.
 

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  • #2
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
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:
 
  • #5
but wouldn't i need to add inertia to the equation?
 
  • #6
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
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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  • #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
 
  • #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.
 

1. What factors affect the velocity of a ball at the top of a loop?

The velocity of a ball at the top of a loop is affected by the initial velocity, the mass of the ball, the radius of the loop, and the force(s) acting on the ball.

2. How does the velocity of a ball change as it goes through a loop?

The velocity of a ball decreases as it goes through a loop due to the force of gravity and any frictional forces acting on the ball. However, if there is enough initial velocity, the ball can maintain a constant velocity at the top of the loop.

3. Can the velocity of a ball at the top of a loop be greater than its initial velocity?

No, the velocity of a ball at the top of a loop can never be greater than its initial velocity. The force of gravity and frictional forces will always cause the ball to slow down as it goes through the loop.

4. How does the mass of the ball affect its velocity at the top of a loop?

The mass of the ball does not have a significant effect on its velocity at the top of a loop. However, a heavier ball may require more initial velocity to maintain a constant velocity at the top of the loop.

5. What is the relationship between the radius of the loop and the velocity of the ball at the top?

The velocity of the ball at the top of a loop is inversely proportional to the radius of the loop. This means that as the radius of the loop increases, the velocity of the ball at the top will decrease.

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