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
Answer
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.
No comments:
Post a Comment