Loop-the-Loop and velocity problem

  • #1

Homework Statement


A bead slides without friction around a loop-the-loop. The bead is released from a height of 10.1 m from the bottom of the loop-the-loop which has a radius 3 m. The acceleration of gravity is 9.8 m/s^2 .

What is its speed at the top of the loop? Answer in units of m/s.

Homework Equations


w=Fd
PE=mgh
KE=1/2mv^2

The Attempt at a Solution


My first thought was to find the PE of the ball at the top, since the KE and the PE are always going to add up to that number, no matter where the ball is on the track. I did mgh = 1/2mv^2 = m(9.8)(10.1) = 98.98m

So the PE = mass times 98.98

then i did 98.98m=1/2mv^2 to get v=14.06982587, then i divided that by two since the ball is at the top of the loop, it's halfway through it i thought to get 7.034912935, which was wrong...

So what am I doing wrong? Any help would be greatly appreciated
 

Answers and Replies

  • #2
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I don't understand why you divided the speed by 2. How does it matter which fraction of the loop the bead made so far? Do you expect it to magically stop at the end, for example?
Your general approach allows to calculate the speed for every height, if you adjust the height difference accordingly.

Please do not drop units in the calculations, they are important tools to check what you are doing and leaving them out is simply wrong.
 
  • #3
haruspex
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then i divided that by two since the ball is at the top of the loop, it's halfway through it i thought
There's no basis for that step.
Forget about the bottom of the loop. Just compare energies at start with energies at top of loop.
 
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  • #4
Just compare energies at start with energies at top of loop.
Ah thank you haruspex i appreciate your help, i got the answer. Thank you for not being sarcastic and rude
 
  • #5
haruspex
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Thank you for not being sarcastic and rude
Oh well, there's always next time.
 
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  • #6
There's no basis for that step.
Forget about the bottom of the loop. Just compare energies at start with energies at top of loop.
What do you mean by comparing the two energies?
 
  • #7
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Something went wrong with the sign of the gR term. A larger radius will lead to a smaller velocity.

Giving full solutions is against the forum rules, but this thread is two years old - just let it rest.

Edit: Merged some posts.
 
Last edited:
  • #8
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sorry

for the mistake in sign
 
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  • #9
haruspex
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What do you mean by comparing the two energies?
PE+KE at release point versus PE+KE at top of loop.
 

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