how is this kind of rolling motion possible?

Question

I was solving this problem :

Suppose you put a sphere in a rough ground with velocity of center of mass $v_{cm}= v_o$ in the positive $x$ axis and with anticlockwise angular momentum $\omega_o$ so velocity due to center of mass and angular velocity at bottom-most point both points in the same direction. What is the relation between $\omega_o$ and $v_o$, so the sphere returns back.

Now friction acts in negative axis and provide both angular retardation and transnational. Only way the sphere can return back if time to make $v_o$ zero is less than time to make$\omega_o$ zero and working out the equations gives an answer which matches with the one given in text i.e $\omega_o < 2v_o/R$.

But how can this be? More importantly if a body starts returning, how can it ever attain pure rolling? ( Which is inevitable since it is being placed in an rough ground)

Answer 1

@ChrisDrost's answer is correct, but we can actually remove the assumption that the friction is constant by considering conservation of angular momentum instead.

If we put our origin at a point along the ground, then there is no net torque on the sphere: The frictional force always points directly towards (or away from) the origin, and the normal force and gravitational force act along the same line and are equal and opposite. Thus, angular momentum about this point is concerned. We also know that the angular momentum of the sphere is given by $$ \vec{L} = m\vec{r} \times \vec{v} + \vec{L}_\text{CM}, $$ which in this case works out to $$ L = m R v + I \omega. $$ A sphere rolling without slipping will have $v = R \omega$, which in this case means that $$ L = (m R^2 + I) \omega = (m R^2 + I) \frac{v}{R}. $$ In particular, if the sphere is to end up rolling without slipping in the negative direction ($v < 0$), then its angular momentum will be negative, and so its initial angular momentum must be negative as well. The initial angular momentum is $m R v_0 - I \omega_0$, so this therefore implies that to get the sphere to return, we must have $\omega_0 > v_0 R/I$. In the case of a solid sphere, this works out to $\omega_0 > \frac{5}{2} v_0/R$.

Answer 2

So this is a phenomenon which is known in billiards as "backspin": you hit a ball off-center and it simultaneously has a motion "forwards" but a spin that imparts a force on the ground to send it "backwards". Trick shots where you induce extreme amounts of backspin by hitting the ball almost vertically downwards are known sometimes as "massé shots", if you want to see some videos of backspin in extreme action.

Such billiard balls do return to a "pure-rolling" state (i.e. rolling without slip) and indeed you can view the curving trajectory of the ball as being due to friction "wanting" to return the ball back to this pure-rolling/no-slip state. Since it's spinning as if it's rolling backwards but traveling as if it's rolling forwards, the force brings these both to some sort of "middle ground": either spinning forwards and traveling forwards or spinning backwards and traveling backwards.

As you note, there is a constant torque $\tau = - \mu ~m~ g ~ R$ on the ball's rotation in addition to the constant force $F = -\mu ~m~ g$ on the ball's forward speed; if $I$ is its moment of inertia then after a time $t$ we see angular velocity and velocity differences $$\begin{array}{cc}\omega = & \omega_0 - \frac{\mu~m~g~R}{I}~ t \\ v = & v_0 - \mu~g~ t\end{array}$$which continue until a time when the rotational equilibrium $ R \omega = - v$ takes over. At this time we can say that the backspin wins if $\omega > 0,$ so $(1 + \frac{mR^2}{I}) ~ \omega > 0,$ so $\omega - \frac{mR}I v > 0$. I've chosen this combination carefully so that $t$ falls out:$$\omega_0 ~-~ \frac{\mu~m~g~R}I ~ t ~-~ \frac{m~R}{I}\left(v_0 ~-~ \mu~ g~ t \right) > 0,$$leaving just$$\omega_0 > \frac{m R}{I} v_0.$$Now the only way I get your quoted result is for $I = \frac 12 m R^2$, which is characteristic of a solid cylinder but not either a solid or hollow ball: in fact the ball would have to have a hollow center of radius $R ~ \frac 14\left(\sqrt{5 + 4 \sqrt{5}} - 1\right)\approx 0.68355 R$ to have this particular moment of inertia, making it about 31.9% empty space by volume rather than the usual values of 0% or 100%.