# Ball rolling in a loop

huskydc
A small ball of radius r = 2.4 cm rolls without slipping down into a loop-the-loop of radius R = 2.5 m. The ball has mass M = 352 gm.

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

Mentor
Hint: What minimum speed must the ball have at the top of the loop in order for it to maintain contact? (It's not zero!)

huskydc
sorry doc, i really have no clue on this...i know i need to find the speed. since here...the ball rolls w/o slipping. thus the velocity is equal to its tangential velocity.

Mentor
Consider the forces on the ball at the moment it reaches the top of the loop. Apply Newton's 2nd law, realizing that the ball is undergoing circular motion.

Homework Helper
The ball will start falling if the Normal force exerts to it is 0 N.
Viet Dao,

huskydc
ok, i got it, thanks!

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marlon
huskydc said:
A small ball of radius r = 2.4 cm rolls without slipping down into a loop-the-loop of radius R = 2.5 m. The ball has mass M = 352 gm.

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)
$$mgh = \frac{mv^2}{2}$$ : this gives v

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

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

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 $1/2I\omega^2$

regards
marlon

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