Question

A bowler throws a bowling ball of radius R = 11.0 cm down the lane with initial speed = 8.50 m/s. The ball is thrown in such a way that it skids for a certain distance before it starts to roll. It is not rotating at all when it first hits the lane, its motion being pure translation. The coefficient of kinetic friction between the ball and the lane is 0.210.
(a) For what length of time does the ball skid? (Hint: As the ball skids, its speed v decreases and its angular speed ω increases; skidding ceases when v = Rω.)
(b) How far down the lane does it skid?
(c) How many revolutions does it make before it starts to roll?
(d) How fast is it moving when it starts to roll?

Answer 1

Answer:

a) 1.18 seconds

b) 8.6 m

c) 5.19 revolutions

d) 6.07 m/s

Explanation:

Step 1: Data given

radius of the ball = 11.0 cm

Initial speed of the ball = 8.50 m/s

The coefficient of kinetic friction between the ball and the lane is 0.210.

(a) For what length of time does the ball skid?

The velocity at time t can be written as v(t) = v0 + at

 ⇒ with v(t) = the velocity at time t

⇒ with v0 : the initial velocity = 8.50 m/s

⇒ with a = the acceleration (in m/s²)

   ⇒The acceleration (negative) due to friction: a = -µg

           ⇒ with µ = 0.210

          ⇒ with g = 9.81 m/s²

v(t) =8.5m/s - 0.21*9.81m/s² * t = 8.5 - 2.06t

Torque τ = Iα = (2m(0.11m)²/5)α = 0.00484m*α

τ = F * r = µm*g*R = 0.21 * M * 9.81m/s² * 0.11m = 0.227m

so α = 0.227m / 0.00484m = 46.9 rad/s²

angular velocity ω(t) = ωo + αt = 0 + 46.9 rad/s² * t

The ball stops sliding when v(t) = ω(t) * r

8.5 - 2.06t  = 46.9*0.11*t = 5.159t

7.219t = 8.5

t = 1.18 seconds

b) How far down the lane does it skid?

s = Vo*t + ½at² = 8.5m/s * 1.18s - ½* 2.06 m/s² * (1.18s)² = 8.6 m

c) How many revolutions does it make before it starts to roll?

The angular acceleration of the ball is:

α =  τ/I

 ⇒ with  τ = the torque experienced by the ball due the frictional force

   ⇒  τ = fk*R

α = fk*R /I

 ⇒ I = 2/5 m*R²

 ⇒ fk = µk*m*g

α = (µk*m*g*R)/(2/5mR²)

α = 5µk*g /2R

The angular displacement of the ball is:

∅ = 1/2αt²

⇒ The ball does not have an initial angular velocity

∅ =1/2*(5µk*g/2)*t²

∅ = 5µkgt²/4R

∅ = (5*0.21*9.81*1.18²)/(4*11.0 *10^-2)

∅ = 32.6 rad

Number of revolutions = 32.6 rad /2π

Number of revolutions = 5.19

(d) How fast is it moving when it starts to roll?

v = Vo + at = 8.5m/s - 2.06m/s² * 1.18s = 6.07 m/s