If a bowling ball is moving with some initial velocity while slipping, how far will it move before it begins to roll once it experiences static friction?

$\ddot{x} = \mu_{kf}g$

And there is also a torque from the kinetic friction on the ball (R = radius of the ball)

$$mg\mu_{kf}R = \frac{2mR^2}{5} \ddot{ \theta} \implies \ddot{ \theta} = \frac{5g\mu_{kf}}{2R}$$

The condition for rolling without slipping is $v = R \omega$ and from the time the ball makes contact with the ground, transversal velocity decreases while angular velocity increases to a point where they are equal. I am not sure what I should do at this point, because everything I try doesn't seem to work.

$$\ddot{x} = v \frac{dv}{dx} = \mu_{kf}g \implies v^2 = (2\mu_{kf}g)x + v_o^2 $$

I don't quite know what to do with this Differential equation that won't involve $\theta$ so that I can use it in the linear equation of motion. I have tried using time, but I don't know how that would help, And the actual angle itself is useless.

$$\ddot{ \theta} = \frac{5g\mu_{kf}}{2R}$$ I can't say $x = R \theta$ because of the slipping

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    $\begingroup$ (Interesting aside): Once it starts rolling-without-slipping, it never ever stops! (unless we include air resistance and/or material deformation) $\endgroup$ Commented Dec 31, 2012 at 6:59

1 Answer 1


Lets say that when your ball first contacts the ground, it has initial velocity $v_0$ and initial angular velocity $\omega_0 = 0$.

You have a constant torque being applied to the ball, so your differential equation is very easy to integrate to get:

$$\dot{\theta} = \omega = \frac{5g\mu}{2R} t + \omega_0$$

For the displacement, go directly with Newton's law, $\ddot{x}=-\mu g$, which also has a constant force and can be easily integrated once to get

$$\dot{x} = v = v_0 - \mu g t$$

From here you should be able to use your $v = \omega R$ condition to find out how long will it take the ball to start rolling without slipping, and once you have that time, integrate displacement once more to get

$$x = v_0 t - \frac{1}{2}\mu g t^2,$$

which will give you the distance traveled entering the time you calculated before.

  • $\begingroup$ Thank you so much. It makes so much sense when you say it $\endgroup$ Commented Dec 31, 2012 at 4:11

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