A bowling ball is given an initial speed $V_o = 10 \text{m/s}$ on an alley such that it initially slides without rolling. The coefficient of friction between the ball and the alley is $\mu = 0.06$. How far does it travel before pure rolling motion occurs?
So I tried solving this problem using the work energy equation, but I'm not too sure if I'm allowed to do this.

So, i considered work done by the ball when it was sliding before rolling which was just kinetic energy and work done by friction and then work done by the ball when it starts rolling which is kinetic + rotational energy. I know initial velocity and i just don't know the final velocity and distance that it was sliding. So, it was my first equation with two unknowns.
$$\frac{mV_i^2}{2} + \mu mgd = Iw^2/2 + \frac{mV_f^2}{2}$$
So, here is question, can i for the second equation assume following:

Just consider the ball right before it starts rolling. So ->

Kinetic energy ($V_i$) = Work done by friction + kinetic energy ( and velocity being Vf which is the same as $V_f$ in my first equation??) Will these velocities be the same?


closed as off-topic by user36790, Kyle Kanos, ACuriousMind, garyp, Floris Dec 15 '15 at 21:17

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  • $\begingroup$ You will need both conservation of energy, and of momentum, in order to solve this. $\endgroup$ – Floris Dec 15 '15 at 9:41
  • $\begingroup$ @Floris There is friction, so energy is not conserved. This is solved using Newtonian dynamics, both linear and rotational. The key is identifying the condition that describes when slipping stops. $\endgroup$ – garyp Dec 15 '15 at 14:56
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    $\begingroup$ @garyp OP correctly writes down the energy equation - KE before = Linear plus rotational KE after plus work done by friction. That is "conservation" in my mind but you can call it something else. But yes there are three equations for energy, linear and angular momentum; and the condition that $b=\omega r$ when slipping becomes rolling. $\endgroup$ – Floris Dec 15 '15 at 17:21
  • $\begingroup$ @Floris That which we call conservation of energy by any other name would smell as sweet ... $\endgroup$ – garyp Dec 15 '15 at 21:13
  • $\begingroup$ Possible duplicate of Rolling bowling ball with slipping $\endgroup$ – Floris Dec 15 '15 at 21:17

Hint: If you denote the velocity and angular velocity of the ball by $v_t$ and $\omega_t$, from the instant the ball starts rolling, they are related by: $$v_t=R\omega_t$$

Construct a free body diagram, and then you can find the net torque and net force on the body. $v_t$ and $\omega_t$ can be obtained from these. Finally, after you find $t$ from the initial relation, this is reduced to a simple kinematics problem.


The acceleration of the ball is $\mu g$ and its angular acceleration (about the center of mass) is $\frac{\mu mgR}{I_{cm}}$.

From these, the velocity and acceleration at a later instant are given by:

$v_t=v_i-at$ and $\omega_t=\alpha t$. (the minus sign because the velocity decreases)

  • $\begingroup$ oh, i tried doing this, but at the end i got negative value for distance. So, when i'm getting negative values for distance or time, does it mean that i did it wrong? Or i can just the absolute value of that? $\endgroup$ – Jack Dec 15 '15 at 5:55
  • $\begingroup$ You shouldn't be getting a negative value. You should check the signs of quantities in the equations you have. Could you add what you've tried to the question? :) $\endgroup$ – Hritik Narayan Dec 15 '15 at 5:56
  • $\begingroup$ Okay, so firstly, there is only force of friction. Hence, F(friction) = ma umg = ma -> a = ug. Now, i used Sum of all torques = Ialpha from here i got Alpha = 1.47/r Then, i used this equation from kinematics w = alphat, by substituting alpha i got w = (1.47*t)/r Then, i used it this w in W = V/r from here i got V = 1.47t. Then i used it in another kinematics equation which is V = Vi+at V = Vi+ugt 1.47*t = 10+ugt t = 0.85 And if i substitute it back to v = 1.47t i got Vf smaller than Vi which then gives me negative distance when i tried finding it. $\endgroup$ – Jack Dec 15 '15 at 6:32
  • $\begingroup$ I've edited the answer to add more. You probably missed a negative sign somewhere. I hope this helps. $\endgroup$ – Hritik Narayan Dec 15 '15 at 6:42
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    $\begingroup$ The frictional force opposes the motion of the body. $\endgroup$ – Hritik Narayan Dec 15 '15 at 7:31

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