Question

A 0.17-kg billiard ball whose radius is 3.3 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 4.2 m/s and the coefficient of kinetic friction between the ball and the billiard table is 0.57.

(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?

Answer 1

Since the impulse passes through the CM,wo = 0 rad/s

then A) t =2*u/(7*mu_k*g) = 2*4.2/(7*0.57*9.81) = 0.214 s

B) S = 12*u^2/(49*mu_k*g) = 12*4.2*4.2/(49*0.57*9.81) = 0.772 m

C) v = 5*u/7 = 5*4.2/7 =3m/s

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derivations of the formulas used

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

A) The ball will start rolling when:
v = rω
u+at = r(ωo+αt)
u+(-μg)t = rωo+(5/2)μgt
t = 2u/7μg

B) The slide length is given by:
s = ut+½at²
s = u(2u/7μg)+½(-μg)(2u/7μg)²
s = 12u²/49μg

C) And the speed when rolling begins is:
v = u+at
v = u+(-μg)(2u/7μg)
v = 5u/7