Problems pertaining to centripetal force

  • Thread starter Gigantron
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  • #1
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Homework Statement


A golf ball with mass m = 0.05 kg is tied to a massless string of length L = 0.28 m. The ball is made to swing in a circle such that the string is horizontal throughout the motion. The ball makes one revolution every t = 2.1 s.

m = 0.05 kg
L = 0.28 m
t = 2.1 s

Homework Equations



Fnet = mv2/r

The Attempt at a Solution



(a) What the magnitude of the tension in the string, Ft in Newtons, while it spins?

First, I made it so that one revolution would occur at the bottom of the circle which is created by the ball swinging, so that it would be a little easier mathematically.

I identified Fnet as the Tension of the string on the ball pulling upward (TSB) minus the weight of the Earth on the ball (WEB). The rearranged equation looks like this:

TSB - WEB = mv2/r

Then I rearranged it to look like this:

TSB = mv2/r + WEB

Since Weight = mg...

TSB = mv2/r + mg

With that, I plugged in my numbers:

TSB = (.05kg)v2/(.14m)+ (.05kg)(9.8 m/s2)

I got the radius by dividing the length by two...though I'm not sure if this is right. Plus, there doesn't seem to be any information about the velocity. How would I go about finding this? Would I use a = ΔV/t?

(b) The string breaks suddenly. How fast does the golf ball fly away, v in m/s?

What accounts for the string breaking? How do I know which direction the string is flying in? Would it matter in this case if the string was flying in a positive direction or a negative direction?
 

Answers and Replies

  • #2
D H
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The ball is made to swing in a circle such that the string is horizontal throughout the motion.

First, I made it so that one revolution would occur at the bottom of the circle which is created by the ball swinging, so that it would be a little easier mathematically.

Read your problem statement again. Bottom of what circle? "The string is horizontal throughout the motion."


Plus, there doesn't seem to be any information about the velocity.
Sure there is. The problem statement tells you the length of the string and the period of one rotation. Assuming uniform circular motion, this gives you the velocity.
 
  • #3
rcgldr
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Assume the ball is sliding on a frictionless horizontal plane (otherwise, the string could not be truly horizontal). The only net force acting on the ball is the centripetal force from the string.
 
  • #4
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Read your problem statement again. Bottom of what circle? "The string is horizontal throughout the motion."
But why would the problem say that it's going in a circle then? I'm really confused.



Sure there is. The problem statement tells you the length of the string and the period of one rotation. Assuming uniform circular motion, this gives you the velocity.

So would I need to do 2∏*(.28) / 2.1 seconds to get the velocity? If this is correct, then would I be able to plug it into the equation I set up for myself successfully?

rcgldr: Assume the ball is sliding on a frictionless horizontal plane (otherwise, the string could not be truly horizontal). The only net force acting on the ball is the centripetal force from the string.
So I should just scratch the "+ mg" I put in my equation if the only thing acting on the ball is the centripetal force from the string?
 
  • #5
rcgldr
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So I should just scratch the "+ mg" I put in my equation if the only thing acting on the ball is the centripetal force from the string?
Yes, the only net force acting on the ball is the centripetal force. (mg is opposed by the upwards force from the frictionless plane so there is no net vertical force).
 
Last edited:
  • #6
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So this is what I ended up doing:

TSB = (.05kg)(.837)2/(.14)

I got the velocity by doing 2∏*(.28m) / 2.1 seconds

I got an answer of .250N exactly...but apparently this is still wrong. What exactly am I not doing here?
 
  • #7
nasu
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What is the 0.14 in the denominator?
 
  • #8
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The length divided by two. I was trying to find the radius ._.
 
  • #9
rcgldr
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The length divided by two. I was trying to find the radius ._.
You can assume the entire length of the string is the radius.
 
  • #10
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Ah, that makes sense. I got .125N as a result and this appears to be the correct answer. Still, I'm not entirely sure about part B for this problem. Should I draw a free body diagram for that part?
 
  • #11
rcgldr
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Still, I'm not entirely sure about part B for this problem. Should I draw a free body diagram for that part?
Assume part B just wants to know the speed of the ball. without regard to direction.
 

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