A small spherical ball of radius r = 1.5 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 2.6 m. The ball has mass M = 358 g.(adsbygoogle = window.adsbygoogle || []).push({});

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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i just calculated the height above zero potential energy (ground) and subtract 2R from it after my calculations.

the initial potential energy, then, is p = mgH.

at the top of the loop, the potential energy is all converted to translational and rotational KE, so...

mgH = 1/2(mv^2 + Iw^2).

the minimum speed needed to barely complete the loop is Sqrt(gR), which, in this case, is ~5.05 m/s.

I = 2/5mR^2 and w = v/R, so Iw^2 = 2/5mv^2

gH = 1/2(v^2 + 2/5v^2) = 1/2v^2(1 + 2/5) = 0.7v^2

so i get H = 1.8216, which is less than the radius of the loop!

this can't be right...

what am i doing wrong?

the equation for the minimum vel is correct, right?

then i should be doing this right...

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# Homework Help: Loop-the-Loop with rotational motion

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