A solid sphere 0.15 m in diameter is released from rest, rolls down a ramp, and

Question

A solid sphere 0.15 m in diameter is released from rest, rolls down a ramp, and drops through a vertical height of h1 = 0.47 m. The ball leaves the bottom of the ramp, which is h2 = 1.33 m above the floor, moving horizontally (Figure 10-27).

(a) What distance d does the ball move in the horizontal direction before landing?
____________________________________ m
(b) How many revolutions does it complete during its fall (i.e., after it rolls off and before it lands)?
____________________________________ rev

Explanation / Answer

Part A)

From conservation of energy

PE = KEr + KEt

Rotational KE for a solid sphere = .5Iw2 where I = (2/5)mr2

Since w = v/r, we get KEr = .5(2/5)(mr2v2/r2) which simplifies to .2mv2

The mgh = .7mv2 (mass cancels)

(9.8)(.47) = .7v2

v = 2.565 m/s

Then to fall 1.33 m, apply d = vot + .5at2 in ther vertical direction

1.33 = 0 + .5(9.8)(t2)

t = .521 sec

Finally d = vt

d = (2.565)(.521)

d = 1.34 m

Part B)

The rotational velocity at the time it leaves the table is

w = v/r

w = 2.565/(.075)

w = 34.2 rad/s

Then apply angular distance = wt

angular distance = 34.2(.521)

angular distance = 17.8 rad

Divide by 2pi for revolutions

The number of revs = 2.84 revolutions.

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