Angular Acceleration during Bowling

[Eggyu]

Homework Statement

A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides on the lane, with initial speed vcm,0 = 6.0 m/s and initial angular speed 0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.34. The kinetic frictional force fk acting on the ball (Fig. 12-35) causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed vcm has decreased enough and angular speed has increased enough, the ball stops sliding and then rolls smoothly.


Figure 12-35

(a) What then is vcm in terms of ?
0.11 m·omega
(b) During the sliding, what is the ball's linear acceleration?
-3.33 m/s2
(c) During the sliding, what is the ball's angular acceleration?
_______rad/s2
(d) How long does the ball slide?
0.514 s
(e) How far does the ball slide?
2.64 m
(f) What is the speed of the ball when smooth rolling begins?
4.29 m/s

Homework Equations

Torque = Rotational Inertia * Angular Acceleration

Tangential Velocity = angular velocity * radius

The Attempt at a Solution

I know I need to use the torque equation but the work I've done have not produced anything logical and my work is now just scattered beyond belief. If you need to see the work for any of part of the problem, i can show you that.

 

[physicsbarrel]

For part (c)

I * alpha = r*(friction coefficient)*mg=(2/5)mr^2*alpha

I hope this helps.

Also, can you show how to do parts (d), (e), and (f)?

 

[Moetasim]

I would approach this problem by first identifying the key variables and equations that are relevant to the situation. In this case, we have a bowling ball with an initial linear and angular velocity, sliding on a lane with a coefficient of kinetic friction. We are asked to find the final speed of the ball and the time and distance it takes for the ball to stop sliding and start rolling smoothly.

To find the final speed (vcm), we can use the equation for tangential velocity: vcm = r*omega, where r is the radius of the ball and omega is the angular velocity. In this case, we are given the radius (11 cm) and the initial angular velocity (0), so we can solve for vcm = 0.

To find the linear acceleration of the ball, we can use Newton's second law: F = ma. In this case, the only force acting on the ball is the kinetic frictional force (fk), which is given by the equation fk = mu*mg, where mu is the coefficient of kinetic friction and mg is the weight of the ball. So, we can rewrite F = ma as fk = ma, and solve for a = fk/m = (mu*mg)/m = mu*g, where g is the acceleration due to gravity. Plugging in the values given in the problem (mu = 0.34, m = mass of the ball, g = 9.8 m/s^2), we can calculate the linear acceleration to be -3.33 m/s^2.

To find the angular acceleration, we can use the equation for torque: Torque = Rotational Inertia * Angular Acceleration. In this case, the torque is caused by the kinetic frictional force, so we can rewrite the equation as fk*r = I*alpha, where r is the radius of the ball, I is the rotational inertia (given by I = (2/5)*m*r^2 for a solid sphere), and alpha is the angular acceleration. Solving for alpha, we get alpha = (fk*r)/(I) = (mu*mg*r)/(I) = (mu*g*r)/(2/5*m*r^2) = (5/2)*(mu*g/r). Plugging in the values given in the problem, we can calculate the angular acceleration to be approximately -0.003 rad/s^2.

To find the time it takes for the ball to stop sliding