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...

Physics problem: Superheroes and sticks

Some time ago, when I was around 17 years old, I have started solving physics problems from the book "Physical Problems" edition 1981 written by Savchenko O. Y. I was solving them one by one consecutively. But I did not finish them at the time...

I still have the book. So, last year, after I broke my leg at a soccer game, I had a bit more time, when I could only read, think or watch television. At that point, I had an idea to resume the problem-solving activity, and still solve some problems occasionally, even though my leg is OK now.

Recently, I was talking with a friend about one of the problems during lunch. That one was about a satellite that was losing height due to air friction (this one is interesting to me since the speed of the satellite is actually increasing). The friend suggested that I could write up my solutions in Latex. At that point, I did not feel like any of the solutions I have come up with was good enough to spend time writing them up. But today, I think, I have a good one. I would like to share it here. Who knows, maybe someone will point me to an even easier and nicer solution.

Here is my interpretation of the problem statement (it is Problem 5.42 in my edition of the book):

Superman and Ironman throw heavy sticks upwards at a pole of the Earth. The Superman's stick fell back after 30 days and the Ironman's stick fell back after 7 days. What is the difference between the initial speeds of the sticks?

Solution It is probably possible to verify that the sticks would go far enough that the approximation $\vec{g}=const$ is not valid anymore. Although I did not do it myself. I have assumed from the start that it is not valid for the given travel times.

From the energy conservation law we can obtain the expression for the speed of a stick, for example, on its way up at a distance $r$ from the Earth center:

\[ e=\frac{E}{m}=\frac{v^2}{2} - \frac{GM}{r} \]

Then, by equating it to the energy at the highest point, we get:

$$ e=\frac{v^2}{2} - \frac{gR^2}{r}=-\frac{gR^2}{x} $$

$x$ is the distance from the center of the Earth to the highest point of the trajectory of a stick. Here I have used the definition of the free-fall acceleration at the Earth surface:

$$ g = \frac{GM}{R^2} $$

$G$ is the gravitational constant, $R$ is the Earth radius.

Eliminating the speed we get

$$ v = \sqrt{\frac{2gR^2}{x}}\cdot\sqrt{\frac{x}{R} - 1} $$

On the way up $v=\frac{dr}{dt}$, so:

$$ \frac{dr}{dt} = \sqrt{\gamma}\cdot\sqrt{\frac{x}{R} - 1}\\ \gamma\equiv\frac{2gR^2}{x} $$

I have integrated the above equation by using the following substitution: $$ \frac{1}{\cos^2{\alpha}} = \frac{x}{R} $$

and obtained the time ($t$) it takes the stick to travel one way as a function of the maximum distance from the centre of the Earth ($x$):

$$ t = \frac{1}{\sqrt{\gamma}}\left( x \cos^{-1}\left(\frac{R}{x}\right) + \sqrt{R(x-R)} \right) $$

I would prefer to have it the other way around...

The fact that the stick speed should be bounded between the first $v_1=\sqrt{gR}$ and the second $v_2=\sqrt{2gR}$ cosmic speeds inspired me to use a search algorithm to solve for $x$. To include the speed at the Earth surface I have used the expressions for the total energy at the surface and at the highest point to construct the following system of equations:

$$ e_0 = \frac{v_0^2}{2} - gR \\ x = -\frac{gR^2}{e_0}\\ t = \frac{1}{\sqrt{\gamma}}\left( x \cos^{-1}\left(\frac{R}{x}\right) + \sqrt{R(x-R)} \right) \\ \gamma\equiv\frac{2gR^2}{x} $$

Finally, I used Python to implement an iterative search algorithm to find solutions of the system above. At the start, the limits of the speed are set to:

$$ vmin_0 = \sqrt{gR} \\ vmax_0 = \sqrt{2gR} $$

then we substitute $v_n = (vmin_n + vmax_n) / 2$ until the calculated $2t_n$ is close enough to the travel times given in the problem statement. Until this happens the following iterations are perfomed (based on the fact that $v$ grows monotonuously with $t$):

if $2t_n > 2t$, then $vmax_{n + 1} = v_n$, else $vmin_{n+1} = v_n$ and the next trial for $v$ is calculated as:

$$ v_{n+1} = (vmin_{n+1} + vmax_{n + 1}) / 2 $$

Below is the implementation of the algorithm:

In conclusion, the solution gives 11156 m/s for the Superman's stick and 11084 m/s for the Ironman's stick. Therefore the difference is around 72 m/s.