# Equivalency of conditions involving angular momentum of a rolling ball hitting a wall

A ball of mass $m$, radius $r$ and a moment of inertia $I = \frac 25 mr^2$ is rolling on the floor without sliding with the linear velocity $v_0$. It hit the wall perpendicularly. Find out the velocity $v_k$ of the ball's receding from the wall after a long time after the collision.

The friction coefficient between the ball and the floor equals $\mu$, whereas the friction coefficient between the wall and the ball is very big. The collisions are infinitesimally short. All collisions are perfectly elastic and do not undergo deformation. Ignore the rolling resistance and the air resistance.

As the collision is very short, the forces acting between the wall and the ball are very big, so we may neglect the gravity, friction between the floor and the ball and the floor's reaction force. Then the angular momentum wrt to the axis of ball's tangency to the wall is conserved. This means $I' \omega' = \mathrm{const}$, where $I'$ is the moment of intertia of a ball wrt to that axis and $\omega'$ the angular velocity wrt to that axis.

But why is it equivalent to the condition that $I\omega + mv_yr = \mathrm{const}$ where $v_y$ is the vertical component of the velocity of the ball?

/edit: the official solution:

The coordinate system is used of axis $x$ perpendicular to the wall and directed left, the axis $y$ is perpendicular to the floor and is directed upwards. Positive angular velocities mean a counterclockwise motion.

As the ball is rolling without sliding, it approaches the wall with linear velocity $v_0$ and angular $\omega_0$.

The collision of with the wall is very short, so the contact force and the reaction force are very big. This means that during the collision we may neglect the gravity, the reaction of the floor and the friction of the ball with the floor. In this situation the torques wrt to the axis of the ball's tangency to the wall equal 0. So the total angular momentum is conserved wrt to that axis $$I\omega + mv_y r = \mathrm{const} (1)$$

Because the wall's friction coefficient is very big, during the collision, the ball will stop sliding wrt to the wall. This means that right after the collision the vertical component of the ball's velocity $v_{y2}$ and is angular velocity $\omega_2$ fulfill the formula $v_{2y} = \omega_2 r$. Taking in account that before the collision $\omega = v_0 /r, v_y = 0$, from the conservation on angular momentum (1) we get $$\omega_2 = \frac {I}{I + mr^2} \frac {v_0} r = \frac 2 7 \frac {v_0} r$$ $$v_{2y} = \omega_2 r = \frac 2 7 {v_0}$$

The wall and the ball are ideally elastic, the total work done by the reaction forces perpendicular to the wall equals zero, so the kinetic energy in the direction of $x$ is conserved, so $v_{2x} = - v_0$

After the collision the ball's motion is a projectile with initial velocity $(v_{2x}, v_{2y}$. The floor and the ball are ideally elastic, so the ball will jump infinitely long, reaching the same maximum height (it has no importance for finding out the final horizontal velocity).

During the collision with the floor we will have friction until we get $v_{x_{konc}} = \omega_{x_{konc}} r$. On the other hand, at each collision with the floor the angular momentum is conserved wrt to the axis of the ball's tangency to the floor $$I \omega + m v_x r = \mathrm {const}$$

Hence $$v_{x_{konc}} = \frac{I \omega_2 + mrv_{2x}}{I+mr^2}r$$

/edit2: It should give $\omega_2 = 2/7 \omega_0$, indeed.

We have $$\frac {dp_x}{dt} = N(t)$$ The friction gives upwards acceleration $$\frac {dp_y}{dt} = fN(t)$$ And diminishes the angular velocity $$I \frac {d\omega}{dt} = -fN(t)r$$

Hence $$I \frac {d\omega}{dt} + r\frac {dp_y}{dt} = 0 ~~~~(*)$$ So after integrating and finding the constants $$mrv_{2y} - mrv_0 = mrv_{2y} = mr^2 \omega_2 + I \omega_2 - I \omega_0 = 0$$ So $$\omega_2 = \frac 2 7 \omega_0$$

In fact, the formula (*) is the formula I have problems with. But why is it in fact the formula for angular momentum conservation for the axis of tangency.

• Is $I$ angular momentum or the moment of inertia in your notation? I think the hint is that you exchange rotation energy for potential energy...and you dont try to conserve the angular momentum...it would never go backwards, as we all estimate. – jaromrax Apr 2 '15 at 12:19
• Yes, yes, yes, lapsus linguae :) It's the moment of inertia. Corrected, thanks! Well, but it's difficult to calculate these differences in energy. The official problem solution uses the angular momentum conservation. – marmistrz Apr 2 '15 at 13:15
• Hi @marmistrz are you still interested in an answer to this question? – rmhleo Oct 17 '17 at 12:00
• @rmhleo TBH after those two years I don't really remember what the problem was about and I don't really have time to investigate it right now. But it may be helpful to someone else. – marmistrz Oct 19 '17 at 19:08

Assuming it is an elastic collision, you will have, right after it, a ball with the same rolling motion (anticlockwise) but translating in the opposite direction at $v_0$. Now there is friction because the ball is "sliding", and the new equilibrium movement will be when both are moving without sliding again at $v_k$. Let me know if you do not know how to solve this last problem. In short, the effect of the collision is only to change the direction of $v_0$. The friction between the wall and the ball should have no effect at all (as the collision is infinitesimally short)

UPDATE: I assume that the ball reverses $v_0$ and friction starts until the ball stops sliding. Thus the final angular speed will be $v_f/r$. We have

$v_f=v_0-at_s=v_0-\mu gt_s$ (1)

where $t_s$ is the time it takes to stop sliding. You can obtain $t_s$ from the torque:

$\tau.t_s=\Delta L=\frac{2}{5} m r^2(\frac{v_f}{r}+\frac{v_0}{r})=\mu mgrt_s$ (2)

from here you obtain $t_s$ and replace in (1) to obtain:

$v_f=\frac{3}{7}v_0$

UPDATE 2: if we accept the explanation that the ball will roll up until it stops sliding, reaching $w_2$, then we need to change, in eq. (2), the initial angular speed in the previous solution from $\frac{v_0}{r}$ to $\frac{2v_0}{7r}$.

In such a case we get:

$v_f=\frac{31}{49}v_0$

• The official solution claims that the ball will have the vertical velocity $v_y = 2/7 v_0$ and the horizontal velocity will only change its sign. It is calculated from the conservation equation and the fact that as the friction coefficient is very big so after some time the ball will stop sliding wrt to the wall. – marmistrz Apr 2 '15 at 16:45
• it doesnt make any sense that the ball has a final vertical speed after a long time (also is not what is asked), plus the horizontal speed must slow down because of the horizontal friction. I got $v_f=3/7v_0$, let me check the results again – user66432 Apr 2 '15 at 17:09
• The official solution claims that the vertical speed will decrease until $v_x = \omega r$. – marmistrz Apr 2 '15 at 17:11
• Is it possible for you to post the official solution? who know what kind of implicit assumption they make? plus what is $\omega$ (I mean as a function of the initial data). i might be wrong, but I have a phD in physics and what you mentioned about the official solution still do not make sense at all to me. – user66432 Apr 2 '15 at 17:16
• Posted ;) (15 chars) – marmistrz Apr 2 '15 at 19:51