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A solid ball with mass $M$ and radius $R$ is placed on a table and given a sharp impulse so that its center of mass initially moves with velocity $v_o$, with no rolling. The ball has a friction coefficient (both kinetic and static) $μ$ with the table. How far does the ball travel before it starts rolling without slipping?

The solution I found starts by setting up a conservation of energy and setting $v = rw$:

$$ \ \frac{1}{2}m v_o^2 = \frac{1}{2}m v^2 + \frac{1}{2}Iw^2 \to v_o^2 = \frac{7}{5}v^2 \quad{(1)}$$

It then goes on to say $W = \Delta K_{rotation}$ and solves for $D$ :

$$ \ \int_{0}^{D} F_{f} dx = μmgD= \frac{1}{2}Iw^2\quad{(2)}$$ There are a couple of things I do not understand about this approach. How does $(1)$ account for the loss of energy due to the friction force which is causing the rotation and the slipping that occurs before it starts rolling purely? Second, how does $(2)$ account for the change in center of mass velocity? Wouldn't $W = \Delta K_{rotation} + \Delta K_{transitional}$ ? I am most likely misunderstanding something and help is greatly appreciated.

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    $\begingroup$ In $(2)$, $u$ should be $\mu$, I think... $\endgroup$
    – Gert
    Aug 24, 2020 at 20:32
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    $\begingroup$ This doesn't sound correct. You can't have conservation of energy in (1) [since the ball slips at first]. I'm not sure what makes you claim (2). I'd use conservation of angular momentum about the initial point of contact to get the "final velocity" with which the ball purely rolls and then start from there. $\endgroup$
    – Vivek
    Aug 24, 2020 at 22:29
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    $\begingroup$ And $w$ should be $\omega$. $\endgroup$
    – Qmechanic
    Aug 25, 2020 at 3:20

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I am not sure about this solution. I would set up the equations of motion as follows.

Translational motion: there is only one force acting on the system, which is dynamic friction of modulus $F_d=\mu mg$. The motion is uniformly decelerated with acceleration $a = \mu g$. The (horizontal component of the) translational velocity then follows the equation $$ v(t) = v_0 - at $$

Rotational motion The dynamical friction acts with a torque of modulus $\mu mg R$ on the system (taking the center of the ball as a pole), then the angular acceleration is given by the rotation dynamic equation $I \alpha = \mu mgR$, which gives $\alpha = 2\mu g/(5R)$. The angular velocity $\omega$ as a function of time reads $$ \omega(t) = \alpha t $$

Pure rolling condition is obtained setting $v(t) = \omega(t) R$, which gives you the time needed to reach pure rolling, which is $t = v_0/(a+\alpha R)$. Now you can use this time to compute the distance with the kinematics law of the translation $D = v_0 t - (1/2)a t^2$.

I think you should account for friction in the energy balance of the problem, so $W=\Delta K_{trasl}+\Delta K_{rot}$, but then you would have two unknown variables: $D$ and the final velocity, so you should use kinematics as stated above in this case.

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  • since it's rolling without slipping the friction (static) won't affect the energy , so yes energy is conserved.

  • here is a good demonstration https://www.youtube.com/watch?v=hxa6jAYA980 about similar case , after applying delta k= w ( transltaion )the forces you have are( only weight because again the friction you have is static while applying delta k= w for rotation you have one torque which comes from friction

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Hi there mister Radek Martinez! Good to see you've joined the game! And with a good question too!

Here's what I think:

The energy loss due to kinetic friction makes the ball move slower, thus the change in linear kinetic energy is equal to the loss in friction-induced loss of energy. Energy is left for linear and rotational velocity.
The force going hand in hand with kinetic friction is equal to the kinetic friction coefficient $\mu$ times the normal force acting on the mass, which is $F_n=Mg$, so:

$$F_{friction}=\mu Mg$$

This means the energy loss until the ball starts will be:

$$\int _0^D\mu Mgdx$$

So now we can write the equation for the conservation of energy:

$$\frac{1}{2}M {{{v_o}_{lin}}}^2=\int _0^D\mu Mgdx +\frac{1}{2}I{\omega}^2+ \frac{1}{2}M v_{lin}^2=\mu MgD+\frac{1}{2}I{\omega}^2+ \frac{1}{2}M v_{lin}^2,$$

where $v_{lin}$ with corresponding energy $\frac{1}{2}M v_{lin}^2$ is the linear motion of the ball when it's starting to move due to static friction and $\frac{1}{2}I{\omega}^2$is (a part) of the initial energy left for letting the ball rotate with static friction.

From this, we can easily extract $D$, the distance at which the ball starts to roll with static friction (all constants are known). I'll leave that for you to do.

I hope this answers the question (implicitly).

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