Question

A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/s. The coefficient of kinetic friction between the sliding ball and the ground is μ = 0.28. Once the ball begins to roll without slipping it moves with a constant velocity down the lane.
1) What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?
2) What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?
3) How long does it take the bowling ball to begin rolling without slipping?
4) How far does the bowling ball slide before it begins to roll without slipping?
5) What is the magnitude of the final velocity?

Answer 1

Answer:

1)  The magnitude of the angular acceleration = 67.92 rad/$s^{2}$

2) Magnitude of the linear acceleration = 2.744 m/$s^{2}$

3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s

4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m

5) the final velocity is 6.21 m/s

Explanation:

the given information :

Bowling mass m = 3.6 kg

Radius = 0.101 m

Initial speed $v_{0}$ = 8.7 m/s

Coefficient of kinetic friction μ = 0.28

1) he magnitude of the angular acceleration of the bowling ball is

F = m a

$F_{g}$  = μ N  ,   N = m g

$F_{g}$  = μ m g

1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:

momen inersia of Bowling ball I = (2/5) m $R^{2}$

torque τ = I α

τ = F R

I α = F R

(2/5) m $R^{2}$  α = μ m g R

α = (5 μ g / 2R) μ g R

  = (5 x 0.28 x 9.8/ 2 x 0.101)

  = 67.92 rad/$s^{2}$

2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane

F = - $F_{g}$ , $F_{k}$ is the force of kinetic friction

m a = - μ m g, remove m

the magnitude of linear accelaration is

a = μ g

  = (0.28) (9.8)

  = 2.744 m/$s^{2}$

3) The bowling ball takes time to begin rolling without slipping:

The linear speed, $v_{t}$ = $v_{0}$ - a t

                            $v_{t}$  =  $v_{0}$ - μ g t

the angular speed, ω = ω0 + α t

                                ω = ω0 + (5  μ g/2R ) t

$v_{t}$ = ω R

$v_{0}$ - μ g t = ω0 R + (5  μ g/2R ) t R

7 μ g t/2 = $v_{0}$ + ω0 R

hence,

t = (2 $v_{0}$ + ω0 R)/  7 μ g

ω0 = 0 (no initial spin), therefore

t = 2 $v_{0}$ / 7 μ g

 = 2 x 8.7 / 7 (0.28) (9.8)

 = 0.906 s

4) How long it takes for the bowling ball to begin rolling without slipping, S

S = $v_{0}$  t - (1/2) a $t^{2}$

  = (8.7) (0.906) - (1/2) (2.744) $0.906^{2}$

  = 6.75 m

5) The final velocity

$v_{t}$ = $v_{0}$ - a t

$v_{t}$ = 8.7 - (2.744) (0.906)

$v_{t}$ = 6.21 m/s