Question

A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides on the lane with initial speed v_com,0 = 11.0 m/s and initial angular speed ω₀ = 0. The coefficient of kinetic friction between the ball and the lane is 0.22. The kinetic frictional force f_k acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed v_com has decreased enough and angular speed ω has increased enough, the ball stops sliding and then rolls smoothly.

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(a) What then is v_cm in terms of ω?
m·ω

(b) During the sliding, what is the ball's linear acceleration?
m/s²

(c) During the sliding, what is the ball's angular acceleration?
rad/s²

(d) How long does the ball slide?
s

(e) How far does the ball slide?
m

(f) What is the speed of the ball when smooth rolling begins?
m/s

Answer 1

Radius of the ball, r = 11 cm = 0.11 m

Initial speed of the ball, vx0 = 11.0 m/s

(a) vcm = r*$\omega$ = 0.11*$\omega$ m/s

(b) The linear a comes from F = m*a

where F = -µ*m*g so -µ*m*g = m*a

so a = -µ*g = -0.22*9.8 = -2.16 m/s^2

(c) The torque is µ*M*g*r = I*α where I = 2/5*M*r^2

so µ*g*r= 2/5*r^2*α = 0.4*r^2*α

so α = µ*g/(0.40*0.11) = 0.22*9.8/(0.40*0.11) = 49.0 rad/s^2

(d) v = v0 + a*t and ω = ωo + α*t = α*t

The ball slides until v = r*ω = r*α*t

So,

r*α*t = 11.0 - 2.16*t

=> (r*α + 2.16)*t = 11.0

=> t = 11.0 / (0.11*49.0 + 2.16)

= 11.0 / 7.55 = 1.46 s

(e) The distance, the ball slides,

x = vxo*t + 1/2*a*t^2

= 11.0 * 1.46 + 1/2*(-2.16)*1.46^2

= 16.06 - 2.30 = 13.76 m (Answer)

(f) Speed of the ball when smooth rolling begins,

v = vo + a*t

= 11.0 - 2.16*1.46

= 7.85 m/s (Answer)