Loop-the-Loop and velocity problem

In summary: You know PE at release point, you can work out KE at release point from that. You know PE at top (zero), you can work out KE at top from that. Equate the two, with v_top as the unknown.
  • #1
GiantSheeps
83
7

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
 
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  • #2
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
GiantSheeps said:
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
haruspex said:
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
GiantSheeps said:
Thank you for not being sarcastic and rude
Oh well, there's always next time.
 
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  • #6
haruspex said:
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
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
sorry

for the mistake in sign
 
Last edited by a moderator:
  • #9
Archy1111111 said:
What do you mean by comparing the two energies?
PE+KE at release point versus PE+KE at top of loop.
 

1. What is the Loop-the-Loop and velocity problem?

The Loop-the-Loop and velocity problem is a physics problem that involves calculating the minimum velocity needed for an object to successfully complete a loop-the-loop or vertical circular motion without falling off the track.

2. What are the key factors that affect the Loop-the-Loop and velocity problem?

The key factors that affect the Loop-the-Loop and velocity problem are the radius of the loop, the mass of the object, the gravitational force, and the height of the starting point.

3. How do you calculate the minimum velocity for a successful loop-the-loop?

The minimum velocity for a successful loop-the-loop can be calculated using the formula v = √(gr(R + r)), where v is the minimum velocity, g is the acceleration due to gravity, r is the radius of the loop, and R is the height of the starting point.

4. Can an object successfully complete a loop-the-loop with any velocity?

No, an object cannot successfully complete a loop-the-loop with any velocity. The velocity must be greater than or equal to the minimum calculated velocity to complete the loop without falling off the track.

5. How does the mass of the object affect the Loop-the-Loop and velocity problem?

The mass of the object affects the Loop-the-Loop and velocity problem because it affects the centripetal force needed to keep the object in circular motion. A larger mass requires a greater velocity to produce the necessary centripetal force for a successful loop-the-loop.

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