Physics: Ring sliding onto a step

This problem took me awhile as well. Mainly, its second part took me the longest because I could not get the same answer as was in the book (Savchenko et al, 1981, problem 6.26). If you have any ideas about the second part of the problem, please leave me your feedback in the comments below the post.

Problem statement

A ring of radius $R$ is sliding with a speed $v$ on a floor and hits a corner of a step of height $h$. Determine the height $H$ to which the ring will bounce in the following 2 cases:

• (a) No friction between the ring and the step.
• (b) Friction is present between the ring and the step, but no sliding.

Note: There is no rotation before the collision.

Link to the latex source of the figures

Solution

The solution is based on energy and angular momentum conservation during the collision. For the energy conservation I have convinced myself that if there is no displacement of the center of mass during the collision and no sliding, that should be OK. On the other hand, angular momentum conservation means that there is no torques acting on the ring during the collision (i.e. $\vec{\tau}_{\rm net} = \vec{0}$ with respect to a selected point).

If we consider torques with respect to point $O$ (see Fig. 1). The torques of reaction of the step corner and of the friction force with the corner are 0. There are the force of gravity and the force from the floor. I assume they cancel each other and the sum of their torques is approximately 0 (this might need more detailed discussion).

In both cases the maximum height of the ring after the collision can be expressed using the vertical component of velocity at the last moment of the collision $v_z$ through the energy conservation:

\begin{equation*} H = \frac{v_z^2}{2g} \tag{1} \end{equation*}

It is convenient to express the velocity just after the collision ($\vec{v}_{ec}$) as a sum of two components:

$$ \vec{v}_{ec}=\vec{v}_1 + \vec{v}_2 $$

where $\vec{v}_2$ is parallel and $\vec{v}_1$ is perpendicular to $OC$ (see Fig. 2).

In both cases $v_z$ (vertical component of the velocity just after the collision) is expressed as (see Fig. 2):

$$ v_z = v_2 \cos\alpha + v_1 \sin\alpha \tag{2} $$

where $\cos\alpha=(R-h)/R$. Therefore, in the following part, I will consider $\alpha$ as a known parameter.

Case (a): no friction

Energy conservation during the collision yields:

$$ v^2=v_1^2 + v_2^2 $$

where $v$ is the ring speed before the collision.

Conservation of the angular momentum (with respect to the point $O$) gives:

$$ v(R-h) = v_1 R \Rightarrow v_1 = v\cos\alpha $$

From the two previous equations $v_2$ can be found as:

$$ v_2 = v \sin\alpha $$

Then by inserting the above expressions for $v_1$ and $v_2$ into (2), we get:

$$ v_z= v_2 \cos\alpha + v_1 \sin\alpha = 2 v \cos\alpha\sin\alpha $$

And by using equation (1) we find the final result for the part (a) of the problem: $$ H_a = \frac{v^2\sin^2 2\alpha}{2g} $$

Case (b): friction, no sliding

Here, the friction force will rotate the ring and we need to take the rotation into account in our energy and angular momentum conservation equations.

Rotational kinetic energy around the center of mass of the ring is $$ K_r = \frac{mR^2\omega^2}{2} $$

Therefore the energy conservation yields:

$$ v^2 = v_1^2 + v_2^2 + \omega^2R^2 $$

Using the non-sliding condition $v_1=\omega R$, we can simplify the above equation as follows:

$$ v^2 = 2v_1^2 + v_2^2 $$

Angular momentum conservation will give us the following relation:

$$ v(R-h) = v_1 R + \omega R^2 = 2 v_1 R \Rightarrow v_1 = \frac{v\cos\alpha}{2} $$

Then using the above two equations $v_2$ can be found as: $$ v_2 = v\left( 1 - \frac{\cos^2\alpha}{2} \right)^{1/2} $$

Finally, using equations (1) and (2), as in the previous part we get the following expression for the maximum height of the ring:

$$ H_b = \frac{v^2\cos^2\alpha}{8g} \left[ \sin\alpha + 2 \left( 1 - \frac{\cos^2\alpha}{2} \right)^{1/2} \right]^2 $$

Below is my attempt to explain why the answer above is different from the one in the problem book.

I think the above answer is the correct answer. Although the expression in the problem book is different. I suspect that the authors of the book made a pen-slip when they were working out $v_2$. If they, by accident, ommited the $1/2$ factor, the expression for $v_2$ would then be:

$$ v_2=v\sin\alpha $$

and from here $v_z$ would be:

$$ v_z = v_2 \cos\alpha + v_1 \sin\alpha = v \sin\alpha\cos\alpha + \frac{1}{2} v \cos\alpha\sin\alpha = \frac{3}{4}v\sin2\alpha $$

Then the above expression for $v_z$ would result in the following: $$ H_b = \frac{9}{16}v^2 \frac{\sin^2 2\alpha}{2g}=\frac{9}{16}H_a $$

Physics problem: Ring and solid ball rolling off a table edge

This post is about another problem from the physcs book by Savchenko et al. You might be interested to check out my previous post in this series which was also inspired by the same problem book.

Let's start with my interpretation of the problem statement. At the $t=0$ we have a ring rolling to the right towards table edge as shown in Fig. 1. The ring rolls without sliding. The table height above the ground is $H$ and the ring radius is $R$. We need to determine how far horizontally from the table edge the ring will land. And finally we have to show how the result will change if the ring is replaced by a solid ball.

Latex source for the figure is available here.

Solution

Let's start by determining the rotation angle at which the interaction between the table and the ring (or a ball) becomes 0 (Fig. 2), i.e. at the moment of take-off. We will further refer to this angle as $\alpha$. From the second Newton's law applied to the projections of the forces on the line $OC$, we get:

$$ -N + mg\cos\alpha = m\frac{v^2}{R} $$

where $N$ - is the force of reaction of the table on the ring, $m$ - mass of the ring, $v$ - the speed of the center of mass of the ring.

At the moment of take-off $N=0$, therefore:

\begin{equation} mg\cos\alpha = m\frac{v^2}{R} \Rightarrow v^2 = gR\cos\alpha \end{equation}

We determine the speed of the ring at the moment of take-off from the energy conservation equation:

\begin{equation} mg(R - R\cos\alpha) = I_O\frac{\omega^2}{2} \end{equation}

where $I_O$ - is the moment of inertia of the object (i.e. ring or ball in our case) around the axis passing through the contact point (O) with the table.

No-slide condition $v=\omega R$ and (1), (2) give:

$$ mgR(1-\cos\alpha)=I_O\frac{v^2}{2R^2}=\frac{I_O}{2R^2}gR\cos\alpha \Rightarrow $$$$ \Rightarrow \cos\alpha = \frac{mgR}{mgR + \frac{I_OgR}{2R^2}} = \frac{1}{1 + \frac{I_O}{2mR^2}} $$

Note that for a ring $I_O = 2mR^2 \Rightarrow \cos\alpha=1/2$. This would be the end of our solution if we were to consider the modified problem statement in a newer edition of the problem book.

Now to find the speed at the moment of take-off, we plug the expression for $\cos\alpha$ into (1) to get:

$$ v^2 = gR\cos\alpha = \frac{gR}{1 + \frac{I_O}{2mR^2}} \Rightarrow v = \left[ \frac{gR}{1 + \frac{I_O}{2mR^2}} \right]^{1/2} $$

Let's now start calculating the horizontal distance travelled by the center of mass of our object from the table edge. It consists of 2 components:

$$ x = x_1 + x_2 $$

where $x_1$, $x_2$ are the distances travelled before and after the take-off, respectively.

From geometry considerations (see Fig. 2):

$$ x_1 = R\sin\alpha $$

Then, since along the horizontal axis $F_x = 0$:

$$ v_x = {\rm const} = v\cos\alpha $$

And therefore, the horizontal distance travelled by the object after take-off is

\begin{equation} x_2 = v_x t = v\cos\alpha\cdot t \end{equation}

where $t$ is the time between take-off and landing of the object.

The vertical distance travelled by the object since take-off:

$$ H - (R-R\cos\alpha) = v\sin\alpha \cdot t + \frac{gt^2}{2} $$

Then, using (3), we get an equation with respect to $x2$:

$$ H - R(1 - \cos\alpha) = x_2\tan\alpha + \frac{g}{2} \frac{x_2^2}{v^2\cos^2\alpha} $$

Solving the above equation for $x_2$ and selecting only positive root, we get:

$$ x_2 = \frac{v^2\sin\alpha\cos\alpha}{g}\left( -1 + \sqrt{1+ \frac{2g}{v^2\sin^2\alpha}(H-R+R\cos\alpha)} \right) $$

Using $v^2 = gR\cos\alpha$:

$$ x_2 = R\cos^2\alpha\sin\alpha\left( -1 + \sqrt{\frac{2}{R\cos\alpha\sin^2\alpha}(H-R+R\cos\alpha) +1}\right) $$

Therefore

\begin{equation} x = x_1 + x_2 = R\sin\alpha\left[ \sin^2\alpha + \cos^2\alpha \sqrt{1 + \frac{2(H-R+R\cos\alpha)}{R\cos\alpha\sin^2\alpha}}\: \right] \end{equation}

The above is a general expression for the horizontal distance the object falls from the table edge.

In case of the ring $\cos\alpha=1/2$, $\sin\alpha = \sqrt{3}/2$, and the expression for the distance is:

$$ x = \frac{R}{8}\left( 3\sqrt{3} + \sqrt{\frac{16H}{R} - 5}\right) $$

The above is the answer to the problem question for the case of a ring, it is the same as in the problem book. I could not quite get to it on my several attempts to solve the problem because I was forgetting to take into account the vertical distance traveled by the center of mass of the object before take-off. Due to this omission, I was considering that it fell the complete height $H$ after take-off whereas, in reality, it is $H - (R-R\cos\alpha)$. And because of this, my answer was slightly different from the one in the problem book. The discrepancy between my answer and the one in the book worried me to the point that I have decided to write my solution up in the hope of getting a comment from a curious reader... But I have discovered the omission, when I was carefully writing the solution for this post, and fixed the problem.

Now let's try to answer the last part of the problem question, i.e. which object (ring or full ball) would fall further from the table edge. For this I was not able to analytically prove which one would be further for different $H/R$ ratios, so I decided to plot $\delta=x_{\rm ring} - x_{\rm ball}$ to see, maybe the answer would depend on it.

%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np

def cosa(mI_factor):
    #mI_factor:
    # ring = 2, solid ball = 7/5 
    return 1. / (1. + mI_factor / 2.)

def sina(mI_factor):
    return (1 - cosa(mI_factor) ** 2) ** 0.5

def dist(ratio, mI_factor, R=1):
    ca = cosa(mI_factor)
    sa = sina(mI_factor)
    
    return R * sa * (sa ** 2 + ca ** 2 * (1 + 2 * (ratio - 1 + ca) / (ca * sa ** 2)) ** 0.5)


# dist(7/8, 2) - 3/8 * (3 ** 0.5 + 1)

def delta(ratio, R=1):
    # dist(ring) - dist(ball) for different ratios
    return dist(ratio, 2, R=R) - dist(ratio, 7. / 5., R=R) 

R = 10
r = np.linspace(6/16, 500, 100) # H/R
d = [delta(ri, R=R) for ri in r]

plt.plot(r, d)
plt.xlabel(r"$H/R$")
l = plt.ylabel(r"$\delta=x_{\rm ring} - x_{\rm ball}$")

From the graph above it is obvious that $\delta=x_{\rm ring} - x_{\rm ball} \leq 0$ for all $H/R$. Therefore, the horizontal distance to the table edge will be greater for the ball. Unfortunately, this is not the same answer as in the problem book, they state the inverse...