Bowling Ball A spherical bowling ball with mass m = 6.50 kg and radius R = 0.680

Question

Bowling Ball
A spherical bowling ball with mass m = 6.50 kg and radius R = 0.680 m is thrown down the lane with an initial   speed of v = 9.00 m/s. The coefficient of kinetic friction between the sliding ball and the ground is = 0.12.    Once the ball begins to roll without slipping it moves with a constant velocity down the lane.

What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?

What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?

How long does it take the bowling ball to begin rolling without slipping?

How far does the bowling ball slide before it begins to roll without slipping?

What is the magnitude of the final velocity?

After the bowling ball begins to roll without slipping, compare the rotational and translational kinetic energy of the bowling ball.

Explanation / Answer

Once the ball starts sliding, the friction will produce torque around the center of mass which will produce an angular acceleration given by

Fr R = I alpha

alpha = uk mg R / (2/5mR^2) = 5 uk g / (2R) = 5*0.12*9.8/(2*0.680) = 4.32 rads/s^2

where Fr=uk mg is the friction, R is the radius, I is the moment of inertia of a sphere, and alpha is the angular acceleration. Notice that the frictional force opposes the initial rotation of the ball and it happens to be against the linear movement of the ball.

Now, the linear acceleration of the center of mass is given by

Fr = m a_cm

a_cm = uk m g / m = uk g = 0.12*9.8 = 1.176 m/s^2

notice that this acceleration is directed against the motion of the ball.

The angular velocity as a function of time is given by

w(t) = wo + alpha t

and the velocity of the center of mass as a function of time is

V(t) = Vo - a_cm t

the rolling condition requires that V = w R; as you can see, the angular velocity increases at the same time the velocity is decreasing. The time tr it takes then to get to this point is given by

w(tr) R = V(tr)

alpha tr R = Vo - a_cm tr

solving for tr we get

tr = Vo / (alpha R + a_cm ) = 9 / ( 4.32*0.68 + 1.176) = 2.188 sec

The distance traveled before rolling is given by

X(tr) = Vo tr - 1/2 a_cm tr^2

X(tr) = 9*2.188 - 1/2*(1.176*(2.188)^2 = 16.88 m

The magnitude of the final velocity is then

V(tr) = 9 - 1.176*2.188 = 6.43 m/s

The rotational and kinetic energies can be calculated as

KE = 1/2 mV ^2 = 1/2*6.5*(6.43)^2 = 134.37 J

RE = 1/2 I w^2 = 1/2 2/5 mR^2 V^2 / R^2 = 2/5*KE = 2/5*134.37 = 53.75 J

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