- #1

huskydc

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How high above the top of the loop must it be released in order that the ball just makes it around the loop?

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- Thread starter huskydc
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- #1

huskydc

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How high above the top of the loop must it be released in order that the ball just makes it around the loop?

- #2

Doc Al

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- #3

huskydc

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- #4

Doc Al

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- #5

VietDao29

Homework Helper

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The ball will start falling if the Normal force exerts to it is 0 N.

Viet Dao,

Viet Dao,

- #6

huskydc

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ok, i got it, thanks!

Last edited:

- #7

marlon

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huskydc said:

How high above the top of the loop must it be released in order that the ball just makes it around the loop?

You do not need to calculate actual forces, you know. Just apply conservation of total energy by splitting up the motion in two parts.

1) begin : the height you start from (h)

2) end : the point where you start the loop (height = 0)

and

1) point where you start the loop (same as 2) above)

2) point at the top of the loop (indeed the normal force is just zero here)

For part one you have (R is the radius of the loop)

[tex]mgh = \frac{mv^2}{2}[/tex] : this gives v

For part two : at the top of the loop, you have [tex]\frac{mv^2}{R} = mg[/tex] : from this you have v' at the top of the loop.

Applying energy conservation for part two yields : [tex]\frac{mv^2}{2} = mg2R + \frac{mv'^2}{2}[/tex]. If you replace the v and v' by their calculated expressions you will get [tex]h = \frac{5R}{2}[/tex]

Remember that this only counts for the ball being a point particle. I am not incorporating rotation here. If you wanna, the total kinetic energy must be expanded with the rotational kinetic energy [itex]1/2I\omega^2[/itex]

regards

marlon

Last edited:

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