A solid ball (1 kg, outer radius = 0.1 m) starts at top of a track at rest with

Question

A solid ball (1 kg, outer radius = 0.1 m) starts at top of a track at rest with its center of mass 20 m above ground level. It rolls down an initial hill up and over a semi-circular hill (radius 14 m), then launches horizontally off another hill of the same height, landing below and continuing forward motion. While moving along the track what fraction of the ball's total kinetic energy is always in rotational form? (In other words, find the ratio of KE_q/KE_) At point A. the top of the loop: (use for parts b-c) How fast is the ball's center of mass (COM) travelling along the track? Show that the ball maintains contact with the track at this point. Be quantitative. Find the ball's flight time after running off the end of the hill at point B.

Explanation / Answer

a.)

Balls mass is m = 1 kg, its radius is r = 0.1 m

If the ball rolls with its center of mass velocity v, then its total kinetic energy is,

KEtot = (1/2)I2 + (1/2)mv2where I = (2/5)mr2 and = v/r

or, KEtot = (1/2)(2/5)mr(v/r)2 +  (1/2)mv2

or, KEtot = (1/5)mv2 + (1/2)mv2 = (7/10)mv2

So the rotational kinetic energy is, KEr = (1/5)mv2

So, KEr/KEtot = [(1/5)mv2]/[(7/10)mv2]

or, KEr/KEtot = 2/7

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b.)

At the start the energy of the ball was,

Ei= PE1 = mgh1

at A, let the speed of the COM be v, then the ball's energy is,

Ef = KE + PE2 = (7/10)mv2 + mgh2

From conservation of energy,

Ei = Ef

mgh1 = (7/10)mv2 + mgh2

or, mg(h1 - h2) = (7/10)mv2

or, g(h1 - h2) = (7/10)v2

or, (9.8 m/s2)(20m - 14m) = (7/10)v2

or, v = 9.2 m/s, is the speed of COM at A.

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c.)

At A, let the normal contact force on the ball by the surface be N, then

mg - N = mv2/r, where r = 14m + 0.1 m = 14.1 m

or, N = mg - mv2/r = m(9.8 - (9.2 m/s)2/14.1m) > 0

So, N > 0, thus the balls remains in contact with the track.

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d.)

At B, the ball will have same speed as at A,

so, v = 9.2 m/s

But this speed is horizontal. The ball is launched off B with zero vertical speed. Let the flight time be t, then,

h = ut + (1/2)gt2, where h = 14 m

or, h = 0 + (1/2)gt2, as u = 0

or, 14 m = (0.5)(9.8 m/s2)t2

or, t = 1.69 s, is the flight time..

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....

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