Circular Motion of a ball Problem

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Homework Help Overview

The problem involves a 60 g ball tied to a 50-cm-long string, swung in a vertical circle. The scenario requires determining where the ball lands after the string is released at the top of the loop, with the center of the circle positioned 150 cm above the ground. The discussion revolves around the minimum speed needed to maintain tension in the string and the subsequent projectile motion of the ball.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of critical velocity and its implications for projectile motion, dividing the problem into horizontal and vertical components. There are attempts to clarify the initial conditions and the equations used, with some questioning the accuracy of their calculations.

Discussion Status

Participants are actively engaging with the problem, pointing out potential errors in calculations and seeking clarification on the problem statement. There is a recognition of mistakes in the critical speed calculation, and some participants are revising their approach based on feedback.

Contextual Notes

There is an ongoing discussion about the correct interpretation of the problem and the necessary calculations, with some confusion regarding the values used in the equations. Participants are also addressing the need for clarity in the problem statement itself.

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Homework Statement


A 60 g ball is tied to the end of a 50-cm-long string and swung in a vertical circle. The center of the circle, as shown in Figure, is 150 cm above the floor. The ball is swung at the minimum speed necessary to make it over the top without the string going slack.

If the string is released at the instant the ball is at the top of the loop, where does the ball hit the ground?

http://session.masteringphysics.com/problemAsset/1000600/5/knight_Figure_07_62.jpg

Homework Equations



Critical velocity = [tex]V{c}[/tex] = [tex]\sqrt{r*g}[/tex]
[tex]\Delta[/tex]d=V{i}*t + (1/2)(a)(t^2)

The Attempt at a Solution


I attempted this problem as a projectile motion problem, and divided the calculations into two parts: Horizontal and vertical

vertical Part

So I used this equation: [tex]\Delta[/tex]d=V{i}*t + (1/2)(a)(t^2) and I know the vertical initial is 0 so I can eliminate V{i}*t of the equation and get: [tex]\Delta[/tex]d=(1/2)(a)(t^2)

2 = (1/2)*(9.8)*(t^2)
2 = 4.9*(t^2)
t^2 = 2 / 49
t = 0.64 seconds

Horizontal Part

The horizontal parts only has 3 variables: V, d, and t (which I know from the vertical part). and V = V{c} = [tex]\sqrt{r*g}[/tex] = 0.59 m/s.

and d = v*t = 0.59*0.64 = 0.3776 m

This doesn't look right. I was wondering if someone can point me in the right direction. Any help is greatly appreciated. Thanks.
 
Last edited:
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Looks ok except for a math error in your critical speed calc.
 
yea I see what you mean. I accidently used the mass instead of the radius.

so it would be v_{c}=[tex]\sqrt{r*g}[/tex] = [tex]\sqrt{0.5*9.8}[/tex] = 6.9

therefore:

d = v*t = (6.9) * (0.64) = 4.4 m/s?

Does this seem right?

Edit: Still not the right answer. Help?
 
What are you asked to find? (I don't see a question in the problem statement.)
 
oops.Sorry about that.

If the string is released at the instant the ball is at the top of the loop, where does the ball hit the ground?
 
cse63146 said:
yea I see what you mean. I accidently used the mass instead of the radius.

so it would be v_{c}=[tex]\sqrt{r*g}[/tex] = [tex]\sqrt{0.5*9.8}[/tex] = 6.9

therefore:

d = v*t = (6.9) * (0.64) = 4.4 m/s?

Does this seem right?

Edit: Still not the right answer. Help?
you've made anoteher math error,
[tex]\sqrt{r*g}[/tex] = [tex]\sqrt{0.5*9.8}= \sqrt4.9 = 2.2m/s[/tex] And then calculate the distance d which wil be in meters.
 
yep

d = v*t = 2.2*0.64 = 1.4 m

Thanks a lot!

still can't figure out why I put down 6.9 instead of 4.9...
 

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