A little rolling sphere

Experiment

The easiest way to evaluate this question is simply to ask your girlfriend to kindly pull a rough pillow case under a softball as it rolls in a straight line along a measuring tape on camera as you listen to elevator music. Then you check that the average velocity for the entire trip is in close agreement with the velocity you measure from the end of the pillow case to the end of the measuring tape, taking care to ensure the ball maintains straight line motion for the entire trip.

video

Using this method I found the entire trip (110.5 cm) to occur in 27 24th's of a second while the trip from the end of the pillow case to the 110.5 cm mark (45 cm) to occur in 11 24th's of a second resulting in the average velocity of 98.2 cm/s for the whole trip and 98.18 cm/s for the second half. To within the accuracy of my experimental setup, which I haven't calculated but I am sure is much more than this difference, these speeds are identical.

Theory

Another, less exciting, way is to use theoretical physics.

Leaving aside the forward velocity of the ball (which will give rise to some interesting patterns with a marker placed on the surface of the steel sphere) let us consider what happens to a ball pulled from rest on a piece of paper.

At the beginning of the pull, the ball has no translational or angular velocity which means that it will slip until its linear velocity is synced to its angular velocity, i.e. when $\displaystyle \displaystyle v=\omega R$ . The interface of the ball and paper is slipping, so the paper grips the ball with kinetic friction, given by $\displaystyle \mu_k mg$ . As long as the two surfaces are slipping, the force of friction pulls the ball along with the paper. During this phase of motion, the ball accelerates according to $\displaystyle \dot{v} = \mu_k g$ . Similarly, the force of friction applies a torque to the ball which provides angular acceleration so that the angular velocity evolves as $\displaystyle I\dot{\omega} = \mu_k g R$ . This phase of acceleration with slipping will proceed until $\displaystyle v=\omega R$ at which point the ball will roll uniformly.

In our example, the ball starts out with no angular velocity along the axis perpendicular to the pull of the paper and no translational velocity along the pull of the paper. The paper is pulled at velocity $\displaystyle v_0$ . We have

$\displaystyle \begin{aligned} \omega(t) &= \omega_0 + \frac{\mu_k g R}{I} t\\ &= \omega_0 + \frac{5\mu_k g}{2R}t \end{aligned}$

and

$\displaystyle v(t) = v_0 + \mu_kgt$

Let us move from the lab frame into the frame of the paper where it is convenient to calculate. This shift leaves the angular velocity unchanged and changes the initial velocity of the ball from zero to $\displaystyle v_0$ (to the left)

Setting $\displaystyle v(t) = \omega(t)R$ we find $\displaystyle t^*$ , the time at which the ball begins to roll uniformly:

$\displaystyle \begin{aligned} v_0 - \mu_kgt &= R\frac{5\mu_kg}{2R}t \\ v_0 &= \mu_kgt\left(1+\frac52\right) \\ t^* &= \frac{2v_0}{7\mu_kg} \end{aligned}$

With the values given in the problem statement, this happens at $\displaystyle t \approx 0.073 \mbox{ sec}$ , well before the $\displaystyle 2 \mbox{ sec}$ pulling window is over. For the rest of the window, the ball rolls uniformly with angular velocity $\displaystyle \omega(t^*) = \frac{5\mu_k g}{2R}\frac{2v_0}{7\mu_kg} = \frac57\frac{v_0}{R}$ counterclockwise. The translational velocity of the ball is given by $\displaystyle -v_0 + \mu_kg \frac{2v_0}{7\mu_kg}$ to the left relative to the paper.

Moving back into the lab frame (where the paper is stationary), this is $\displaystyle \mu_kg \frac{2v_0}{7\mu_kg} = \frac27 v_0$ to the right. Concretely, the ball moves with velocity $\displaystyle v=\frac27 v_0$ to the right, relative to the ground and rotates with angular velocity $\displaystyle \frac57\frac{v_0}{R}$ counterclockwise.

This proceeds until the two second mark at which point the paper stops and the ball starts to skid across the paper to the right while rotating counterclockwise. At this point we have the same problem as before in which the paper grips the ball with kinetic friction. This will proceed until the linear velocity and the angular velocity are synced.

We have $\displaystyle v_0 = \frac27 v_0$ and $\displaystyle \omega_0 = \frac57\frac{v_0}{R}$ .

These values evolve according to $\begin{aligned}\displaystyle \omega(t) &= \omega_0 - \frac{\mu_k g R}{I} t \\ v(t) &= v_0 - \mu_kgt\end{aligned}$

I.e.

$\begin{aligned}\displaystyle \omega(t) &= \frac57 \omega_0 - \frac{\mu_k g R}{I} t \\ v(t) &= \frac27 v_0 - \mu_kgt\end{aligned}$

Solving $\displaystyle v(t) = \omega(t)R$ we find that these are matched at $\displaystyle t^* = \frac27\frac{v_0}{\mu_kg}$ . However, this is the point where both quantities are precisely zero, which implies that the whole tug-of-war ends as it started off… at rest.

This shows that the effect of the paper pulling is simply to spin the ball up to a uniform rotation and translation which dissipate perfectly at the end of pulling. Nothing about the process of the paper pulling does anything to affect the energy of the original rolling motion and so, that energy is unchanged and $v_i/v_f=1$ .

Inference

Yet another way is to perform a meta-analysis of the problem on its own. This was actually my original method of solution.

It is very hard for me to believe that the paper pulling can contribute or dissipate any overall energy to/from the ball in such a way that no matter what the value of the starting energy, the spinning would have the same relative effect on the overall energy of the ball, i.e. on the value of $\displaystyle v_i/v_f$ .

For this reason, I assigned the belief that for a problem such as this, the actual value of the initial velocity would be important given that the paper affects the final velocity to be $\displaystyle \mbox{P}(v_i \mbox{ important}|\mbox{ paper does something}) = 0.95$ . Similarly, I assigned $\displaystyle \mbox{P}(v_i \mbox{ un-important}|\mbox{ paper does nothing}) = 0.99$ .

Before thinking about the problem I was indifferent to the $\displaystyle v_i$ mattering so $\displaystyle \mbox{P}(v_i \mbox{ important}) = 0.5$ .

I wouldn't be blown away if the paper did something and $\displaystyle v_i$ wasn't shown so

$\displaystyle \mbox{P}(\mbox{paper does something}|v_i\mbox{ not shown}) = 0.1$

but I wouldn't be surprised at all if $\displaystyle v_i$ was shown and the paper didn't do a thing... so

$\displaystyle \mbox{P}(\mbox{paper does something}|v_i\mbox{ not shown}) = 0.5$ .

My belief that the paper does something should then be given by

$\displaystyle \displaystyle \mbox{P}(\mbox{paper does something}) = \frac{\mbox{P}(\mbox{ paper does something}|\mbox{ }v_i\mbox{ not shown})\mbox{P}(v_i\mbox{ not shown})}{\mbox{P}(v_i \mbox{ not shown}|\mbox{ }v_i\mbox{ paper does something})} = \frac{0.1\times0.5}{0.95}$

and that the paper does nothing $\displaystyle \displaystyle \frac{0.5\times0.5}{0.05}$ giving the normalized belief of $\displaystyle 1.04\%$ that the paper matters and $\displaystyle 98.96\%$ that the paper doesn't matter. Given this very strong belief, I inferred that the paper left the ball's velocity unchanged and thus $\displaystyle v_i/v_f = 1$ .

Let us say motion's plane is $X-Y$ and initially, the ball's velocity( $\vec{v_{0}}$ ) was along $X$ -axis ( $\hat{i}$ ), Clearly, $\vec{\omega_{0}}$ was along $\hat{j}$ . Clearly, $v_{0} = \omega_{0}R$ , $\vec{R}$ is from center to bottom point.

$\vec{f} = \mu_{k} mg \hat{j}$

Hence, $\vec{a} = \mu_{k} g \hat{j}$

Similarly, taking moment , angular acceleration = $\vec{\alpha} = \frac{\vec{R} \times \vec{f}}{\frac{2}{5}mR^2} = \frac{5}{2}\frac{\mu_{k}g}{R} \hat{i}$ ,

Thus, $\vec{v_{cm}} = v_{0} \hat{i} + \mu_{k}gt \hat{j}$ ,

$\vec{\omega} = \frac{5}{2}\frac{\mu_{k}gt}{R} \hat{i} + \omega_{0} \hat{j}$

Now, $\vec{v_{bot.}} = \vec{v_{cm}} + \vec{w} \times \vec{R}$

= $\frac{7}{2} \mu_{k} gt \hat{j}$ ,

When friction ceases , $\vec{v_{bot}} = \vec{v_{paper}} = 1 \hat{j}$

Hence, after $t = \frac{2}{7 \mu_{k} g} \approx 0.073 s$ , friction is 0., when $v_{\hat{j}} = \frac{2}{7} m/s, \omega_{\hat{i}} = \frac{5}{7R}$

Now, when paper is stopped, $v_{paper} = 0$ , $v_{bot} = 1 \hat{j}$ , hence, now friction would be along $-\hat{j}$ , again having magnitude $\mu_{k} mg$ .

Now, from above, we saw that $\vec{v_{bot}}$ will be changed only along $\hat{j}$ as the force and torque by friction is tending to change only $\hat{i}$ component of $\vec{\omega}$ .Hence, we can reduce our math to scalar where $\omega$ and $v$ have fixed directions, we won't be considering other directions because along them , changes aren't taking place.In the following , the component of $a$ and $v$ discussed is along $\pm \hat{j}$ and $\omega$ and $\alpha$ in $\pm \hat{i}$ .Let us change $t=0$ to the point when paper is stopped.

As done before, we can easily make equations,(taking $\omega_{\hat{i}}$ and $v_{\hat{j}}$ +ve)

$a = -\mu_{k}g$ ,

$\alpha = -\frac{5}{2} \frac{\mu_{k} g}{R}$

$v = \frac{2}{7} - \mu_{k}gt$

$\omega = \frac{5}{7R} - \frac{5}{2} \frac{\mu_{k} g}{R}$

$v_{bot.} = v + \omega R = 0 \Rightarrow t_{0} = \frac{2}{7 \mu_{k}g}$ is the time when pure roll. begins. Now , $v_{t_{0}} = 0$ , on substituting values.

Hence , finally, $\vec{v} = v_{0} \hat{i}$ , ratio = $\frac{v_{0}}{v_{0}} = \boxed{1}$

WHEN THE SPHERE ROLLS ONTO THE PAPER

THE PAPER IS PULLED FROM UNDER IT IMMEDIATELY......THIS SUGGESTS THAT THERE WILL BE

SLIPPING BETWEEN THE PAPER AND THE SPHERE IN THE DIRECTION IN WHICH IT IS PULLED

LET THE VELOCITY ACQUIRED BY THE SPHERE BE $v_{cm}$ AND THE ANGULAR VELOCITY BE $\omega$

NOW...LINEAR IMPULSE PROVIDED BY FRICTION

$p=mv_{cm} \Rightarrow \mu_{k}mgt=mv_{cm}$

$\Rightarrow v_{cm}=8m/s$

ANGULAR IMPULSE PROVIDED BY FRICTION

$J=I_{cm}\omega \Rightarrow \mu_{k}mgtR=I_{cm}\omega$

$\Rightarrow \omega=2000rad/s$

ONCE AGAIN SEE THAT THE SPHERE SLIPS IN THE DIRECTION IN WHICH THE PAPER HAS BEEN PULLED WHILE IT ROLLS IN THE OTHER DIRECTION

HENCE FRICTION WILL ACT ONLY IN THE DIRECTION IN WHICH THE SPHERE SLIPS

ie-THE DIRECTION IN WHICH THE PAPER WAS PULLED

ALSO OBSERVE THAT THE DIRECTIONS OF $v_{cm}$ and $\omega$ ARE OPPOSITE

NOW...........ANGULAR MOMENTUM ABOUT THE POINT OF CONTACT OF SPHERE WITH THE

PAPER WILL ALWAYS BE CONSERVED (AS NET TORQUE ABOUT THAT POINT IS ZERO )

$\Rightarrow L_{i}=L_{f}$

$L_{i}=I_{cm}\omega+m(-v_{cm})R$

$I_{cm}=\frac {2}{5}mR^2$

$\Rightarrow L_{i}=0 \Rightarrow L_{f}=0$

THIS IS ONLY POSSIBLE IF $v_{cmfinal}=R\omega_{final}=0$

THIS CLEARLY MEANS THAT AFTER A LONG TIME THE SPHERE HAS STOPPED ALL MOTION IN THE DIRECTION IN WHICH THE PAPER WAS PULLED

WHILE IT IS STILL IN A STATE OF PURE ROLLING IN ITS ORIGINAL DIRECTION OF MOTION

$\Rightarrow \boxed{v_{i}=v_{f}}$

Let $\mathbf{i},\mathbf{j},\mathbf{k}$ be a right-handed orthonormal triad of vectors, with $\mathbf{i},\mathbf{j}$ both horizontal, with $\mathbf{i}$ pointing in the initial direction of travel of the ball, and $\mathbf{k}$ pointing vertically upwards. The initial velocity of the centre of mass of the bass is $\mathbf{v} \,=\, v_i\mathbf{i}$ , and the initial angular velocity of the ball is $\mathbf{\omega} \,=\, \omega_i\mathbf{j}$ , where $v_i \,=\, a\omega_i$ , where $a$ is the radius of the ball. This last rolling condition ensures that the point on the ball in contact with the table is stationary, since it has velocity $\mathbf{v}_C \; = \; \mathbf{v} + \mathbf{\omega} \wedge (-a\mathbf{k}) \; = \; v_i\mathbf{i} - a\omega_i\mathbf{j}\wedge \mathbf{k} \; = \; \mathbf{0}$

When the paper starts to move, and while the ball continues to slide on the paper, the ball is subject to a friction force $\mu_k mg\mathbf{j}$ acting at the point of contact of the ball with the paper, where $m$ is the mass of the ball, and hence we have the two equations of motion $m\mathbf{\dot{v}} \; = \; \mu_k mg\mathbf{j} \qquad I\mathbf{\dot{\omega}} \; = \; -a\mathbf{k} \wedge \mu_k mg\mathbf{j} \; = \; a\mu_kmg\mathbf{i}$ where $I = \tfrac25ma^2$ is the moment of inertia of the ball about its centre. Thus $\mathbf{\dot{v_C}} \; = \; \mathbf{\dot{v}} + \mathbf{\dot{\omega}} \wedge (-a\mathbf{k}) \; = \; \mu_k g \big(1 + \tfrac{ma^2}{I}\big)\mathbf{j} \; = \; \tfrac72\mu_kg \mathbf{j}$ This shows us that the motion of the paper relative to the point of contact with the ball is always in the direction of the vector $\mathbf{j}$ , and so the friction force continues to point along the vector $\mathbf{j}$ . Thus we deduce that $\mathbf{v_C} \; = \; \tfrac72 \mu_k g t\mathbf{j} \; = \; 1.4gt \mathbf{j}$ and this model is valid for $0 \le t \le \tfrac{1}{1.4g} < 2$ , at which time $\mathbf{v_C} = \mathbf{j}$ , with $\mathbf{v} \; = \; v_i\mathbf{i} + \tfrac27\mathbf{j} \qquad \mathbf{\omega} \; = \; \tfrac{5}{7a}\mathbf{i} + \omega_i\mathbf{j}$ The ball is now simply rolling on the moving paper. The key point to note is that the sliding stops (well) before the $2$ seconds of paper movement are up.

When the paper stops moving, this is effectively applies the same friction forces as before, but in the opposite direction, so that the ball is now subject to a sideways force of $-\mu_k mg\mathbf{j}$ , acting at the point of contact with the table, and the ball will slide on the table top until $\mathbf{v_C}$ returns from $\mathbf{j}$ to $\mathbf{0}$ . Considering the above calculations, this means that a total of $\tfrac27\mathbf{j}$ will be subtracted from the velocity $\mathbf{v}$ , and so the ball's velocity will change from $v_i\mathbf{i} + \tfrac27\mathbf{j}$ back to $v_i\mathbf{i}$ , so that $v_f = v_i$ , and hence $\frac{v_f}{v_i} = 1$ .

As initially the sphere is rolling the point of contact is not moving wrt the paper so the frictional force doesnt effect the initial anqular velocity component in its direction. there will be an additional angular velocity in the perpendicular direction as friction force is always perpendicular to the initial direction of velocity of center of mass of the sphere . We can take this situation as super imposing the sphere rolling in one direction and slipping and rolling in the direction in which the paper is pulled. The effect of the friction will cause the sphere to roll after some time in both directons. and after 2 sec friction force will act in opposite direction to the initial direction of friction which will again cause slipping untill the gained angular velocity and velocity of center of mass become zero in a direction where the paper is pulled. so finally the sphere moves in the same direction as before rolling with the same velocity of the center of mass.* Even if the sphere doesnt start rolling in the direction perpendicular to its initial velocity before 2 seconds still after two seconds the effect of friction will be opposite irrispective of whether the ball rolls or not.* so the ratio of final to initial velocity is independent of the time the paper is pulled so no calculation is required