An Unconventional Game of Billiard

First, let's forget and change the backspin $w_{o}$ to $w_{s}$ for interpretation purposes.

For the cue ball, we need to guarantee the conservation of momentum in the exact moment after it is hitten by the cue and then it is released from contact with it at a speed of $v_{o}$ and a backspin of $w_{s}$ . Thus, in that instant we will proceed with the total kinetic energy of the ball as:

$K=KE_{t} + KE_{r}$

$K=\frac{1}{2}mv_{o}^2+\frac{1}{2}Iw_{s}^2$

From here, we have two approaches, but I'll decide to work in a translational movement basis, so I'll turn the rotational energy into its translational equivalent using $v_{s}=w_{s}r$ and the moment of inertia of a sphere $\frac{2}{5}mr^2$ . Then,

$K=\frac{1}{2}mv_{o}^2+\frac{1}{2}(\frac{2}{5}mr^2)(\frac{v_{s}}{r})^2$

$K=\frac{1}{2}mv_{o}^2+\frac{1}{5}mv_{s}^2$ ,

and we derivate the energy equation with respect to the speed to get the momentum,

$\frac{dK}{dv}=mv_{o}+\frac{2}{5}mv_{s}$ ,

Because the velocities are in opposite directions, to achieve the return of the cue ball, the condition is that,

$v_{0}<\frac{2}{5}v_{s}$

That's why:

$5v_{0}<2w_{s}r$

There are two approaches to this problem that I could find: impulse/momentum principles and basic kinematics. My solution will focus on the former, but I encourage you to post your own solutions as well!

I'm assuming a 2D environment here where the cue ball rolls in the POSITIVE x direction. Impulse momentum principle in linear direction states:

$\int_{\Delta t} \sum F_x dt = \Delta(mv)$

Free body diagram reveals the friction force $f_k$ in the negative x direction, so: $\sum F_x = -f_k = -\mu_kN = -\mu_kmg$ . Furthermore, the initial velocity of the cue ball is $v_0$ and its final velocity when it rolls without slip is $v$ , so we get:

$\int_{\Delta t} -\mu_kmg dt = mv - mv_0$

$-\mu_kmg \Delta t = m(v - v_0)$

$\mu_kg \Delta t = v_0 - v$ (1)

Now for angular momentum principle, I'm assuming clockwise direction to be positive, so the initial angular velocity is $-\omega_0$ . The final angular velocity when it rolls without slip is $\omega$ and the sum of moments on the ball relative to its center of mass is simply $f_kr = \mu_kmgr$ . Angular momentum principle holds:

$\int_{\Delta t} \sum M dt = \Delta(I\omega)$

The moment of inertia of a sphere is $\frac{2}{5} mr^2$ therefore we get:

$\int_{\Delta t} \mu_kmgr dt = I\omega - I(-\omega_0)$

$\mu_kmgr \Delta t =\frac{2}{5} mr^2(\omega + \omega_0)$

$\mu_kg \Delta t =\frac{2}{5} r(\omega + \omega_0)$ (2)

Notice the similar term $\mu_kg \Delta t$ in both equations (1) and (2) , we can equate them to obtain:

$\frac{2}{5} r(\omega + \omega_0) = v_0 - v$

I made the assertion that the ball will start rolling without slip after a certain amount of time, this implies $v = r\omega$ or $\omega = \frac{v}{r}$ so we can plug this in to get:

$\frac{2}{5} r(\frac{v}{r} + \omega_0) = v_0 - v$

$\frac{2}{5}v + \frac{2}{5}r\omega_0 = v_0 - v$

$\frac{7}{5}v = v_0 - \frac{2}{5}r \omega_0$

$v = \frac{5v_0 - 2r \omega_0}{7}$

We want the cue ball to always come back, so we need to impose $v$ to be negative, that is, travelling in the negative x direction. So the condition is:

$5v_0 < 2r \omega_0$

Notice that if $5v_0 > 2r \omega_0$ then the no slip velocity will be positive and the ball will always roll away.

There is a friction force $F$ acting at the bottom of the ball that pulls it backward. The equation of motion for the center of mass is $m\frac{dv}{dt}=F$ . The equation of motion for the rotation around the center of mass is $\frac{2}{5}mr^2\frac{d\omega}{dt}=Fr$ , where $r$ is the radius, $\frac{2}{5}mr^2$ is the moment of inertia and $Fr$ is the torque. Eliminating $F$ we get $\frac{2}{5}mr\frac{d\omega}{dt}=m\frac{dv}{dt}$ , or

$2r\frac{d\omega}{dt}=5\frac{dv}{dt}$

This equation sets a relationship between the rate of change of the linear and rotational velocities. For the ball to have a backward velocity we need $5v_0<2\omega_0 r$

This question does not require lengthy calculations. Just remember the angular momentum is conserved in respect to the ground (where the frictional force is being applied, so the torque is zero). And for the ball to return, its total angular momentum must be in the same direction with the angular momentum generated by backspin $\omega_0$ and opposite to angular momentum generated by linear velocity $v_0$ . So the equation is:

$I\omega_0 > rmv_0$

where $I$ is the moment of inertia of the ball. For a sphere of mass $m$ and radius $r$ , $I=\frac{2}{5}mr^2$ . So,

$\frac{2}{5}mr^2\omega_0 > rmv_0$

$2r\omega_0>5v_0$

Conservation of angular momentum.

Here we have a little tricky skill about using this:

Lets choose one point (anyone) on the table as the reference axis(exactly on the point),and write down the angular momentum of the cue ball at $t=0$ :

As the friction force points toward the axis, $M=0$

So we can use the conservation of angular momentum.

$L=mv_0r-\frac{2}{5}mω_0r^2=const$

Assuming that the direction poingting into the screen is the positive direction.

if the cue ball moves toward the hole, the angular momentum of it shold be negative.(poingting our of the screen)

$mv_0r<\frac{2}{5}mω_0r^2$

$5v_0<2ω_0r$

【I'm really sorry if I didn't express it in a formal way because I'm not sure about some words......English is not my mother tongue(￣▽￣)／】

Note: From t=0 to t=t', the ball is slipping forward (not rolling), hence friction acts backward

Let the friction force to be $f=ma$ where $a$ denotes acceleration. Then, we calculate angular acceleration: $\begin{aligned}fr&=I\alpha\\&=\frac25mr^2\alpha\\\alpha&=\frac{5f}{2mr}\end{aligned}$

Then, we use two elementary relations about velocity and angular velocity: $\begin{cases}v=v_0+at\\\omega=\omega_0+\alpha t\end{cases}$ Plugin expressions $a=\frac fm$ and $\alpha=\frac{5f}{2mr}$ we obtain their relation: $r\alpha=\frac52a$ In addition, if the cue ball successfully trace back and always contact with the surface, it should have $v<0$ and $v=r\omega$ if the initial velocity $v_0$ is assumed to be positive.

By the knowledge of these concepts, we can rewrite the expressions as $\begin{cases}5v=5v_0+5at\\2v=2r\omega_0+5at\end{cases}$ Subtract to get $3v=5v_0-2r\omega_0<0\rightarrow\boxed{5v_0<2r\omega_0}$ .

I tried to solve this with conservation of energy. Kinetic energy of forward movement has to be lower than kinetic energy of backward spin in order to the ball go back, so:

$KEm < KEs$

$\frac{1}{2}mv^2 < \frac{1}{2}I\omega^2$

$mv^2 < \frac{2}{5}mr^2\omega^2$

$5v^2 < 2r^2 \omega^2$

I would be very grateful if someone told me where i made mistake.

I'll will again post very simple cheating solution since everybody above me has got the right ones already =)

1)v<rw => v<rw

2)v<2rw => v<2rw

3)2v<rw => v<0.5rw

4)5v<2rw => v<0.4rw

If option 1 is the right answer, then 3 and 4 must be right as well, since if v<0.5rw or v<0.4rw is true, v<rw is definitely true, so 3 and 4 are also sufficient condition.

By the same logic, if option 2 is true, then 1,3 and 4 are all true. All options guarantee the ball will return.

Lastly, if option 3 is true, option 4 is also true.

Since there is only one correct answer, the only option that doesn't result in another option being true is option 4.

1st apply newtons 2nd law for translatory motion V=v_0-ft/m So time when ball stops for the first time is ; t=(mv0)/f Now newtons 2nd law for rotational motion ; w=w0-5ft/2mr=>w=w0-(5v0)/2r As ,w>0; 5v0<2w0 r

What I did was understand that more backspin and more velocity will probably ensure the trick shot, so I picked the last option.

With physics problems like this I usually just throw numbers and formulas around and see what happens.

Momentum has to be conserved. So if we want the linear momentum to balance the rotational momentum we need formulas for each.

I know the linear momentum formula is $mv$ and I can look up angular momentum for a sphere. Angular momentum = moment of inertia times angular velocity becomes $\frac{2}{5}mr^{2}*\omega$ (At this point only one answer makes sense because of the 5. Nothing's going to cancel that, but let's keep going.)

If I set them equal then the $m$ 's cancel and I can write $5v=2r^{2}\omega$ . Obviously, we need a little extra spin, so the angular side is greater.

I picked the right one, but you can see I did something a little wrong because my answer's got $r^{2}$ it should just have $r$ .