Cue Billiard Cue Stick
Cue Billiard Cue Stick
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 58 Inch 2 Piece Fiberglass Billiard Pool Cue Stick 21 ounce US $24.95
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 58 2 Pce Hardwood Canadian Maple Pool Cue Billiard Stick 20 Oz W Steel Joint US $27.95
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 58 2 Piece Fiberglass Pool Cue Billiard Stick 20 oz US $27.95
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 57 2 Piece Maple House Cue Pool Billiard Stick 18 oz US $15.95
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 Black 58 Hardwood Canadian Maple Pool Cue Billiard Stick 20 oz W Steel Joint US $27.95
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A 0.17-kg billiard ball whose radius is 3.4 cm is given a sharp blow by a cue stick.?
A 0.17-kg billiard ball whose radius is 3.4 cm is given a sharp blow by a cue stick. The applied force is horizontal and the line of action of the force passes through the center of the ball. The speed of the ball just after the blow is 3.9 m/s and the coefficient of kinetic friction between the ball and the billiard table is 0.61.
(a) How long does the ball slide before it begins to roll without slipping?
(b) How far does it slide?
(c) What is its speed once it begins rolling without slipping?
Thank you!
Given m=0.17kg, r=3.4cm, u=3.9m/s, μ=0.61, ωo=0
The ball's linear deceleration due to friction is:
a = F/m = (-μmg)/m = -μg
The moment of inertia of a solid ball is: I=2mr²/5
So the ball's angular acceleration upon rolling is:
α = rF/I
α = r(μmg)/(2mr²/5) = (5/2)μg/r
The ball will start rolling when:
v = rω
u+at = r(ωo+αt)
u+(-μg)t = rωo+(5/2)μgt
t = 2u/7μg
The slide length is given by:
s = ut+½at²
s = u(2u/7μg)+½(-μg)(2u/7μg)²
s = 12u²/49μg
And the speed when rolling begins is:
v = u+at
v = u+(-μg)(2u/7μg)
v = 5u/7
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