Physics Challenge April 2020

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Summary:

This is an experiment if we could establish a little physics competition.
Classical Physics, Special Relativity Theory

Main Question or Discussion Point

Problem 1 (@wrobel )
(solved by @TSny )

There is a perfectly rough horizontal table. This table is pretty wide (actually it is a plane) and it rotates about some vertical axis. Angular velocity is a given constant: ##\Omega\ne 0##. Somebody throws a homogeneous ball on the table. The ball has a mass ##m## and a radius ##r##.
The ball begins to roll on the table without slipping. Initially (##t=0##) a vector ##\boldsymbol{OP}## (see the picture) is known; initial angular velocity of the ball and initial velocity of its center of mass ##S## are also known vectors.
1) Prove that the center of the ball describes a circle relative to the lab frame;
2) Find the radius of this circle.
1587049401779.png





Problem 2 (@wrobel )
(solved by @TSny , @Chestermiller )

There are four homogeneous rods. Each rod has mass ##m## and length ##b##. The ends of the rods are connected by frictionless hinges such that the rods form a rhombic frame (see the picture). This frame is shaped as square ##ABCD## and put at rest on a smooth horizontal table. Then one applies a force ##F## to the hinge ##A## along the diagonal #AC#. Find acceleration of the point ##C## right after the force has been applied.
1587049421989.png





Problem 3 (@PeroK )
(solved by @Gaussian97 )

a) Suppose a particle has three-velocity ##\vec{u} = (u_x, u_y, u_z)## in frame ##S##, which is moving with velocity ##(v, 0, 0)## relative to frame ##S'##. Show that the gamma factor of the particle in frame ##S'## is given by:
$$\gamma'_u = \gamma_v \gamma_u(1 + \frac{vu_x}{c^2})$$

b) Imagine a rig, consisting of two vertical poles, a distance ##L## apart. Two synchronised clocks are mounted on the poles and moving upwards at speed ##u##, relative to the rig, remaining horizontally aligned with respect to the rig.

Imagine a frame of reference in which the whole rig is moving horizontally at speed ##v##. By what amount are the clocks out of synchronisation in this frame?

c) Imagine instead that the clocks are aligned horizontally in the frame in which the whole rig is moving horizontally, as above. Note that the clocks are at rest relative to each other and assume they are synchronised in their own rest frame. By what amount are they out of synchronisation in the frame in which the rig is moving?
 
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Answers and Replies

  • #2
Spinnor
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I tried to perform an experiment to show the conjectured circular orbit in the 1st problem above. My experiment can be seen here,


The results were suggestive but all my orbits were outward spirals. Question, we are told the sphere is homogenous, I assume that means it is a uniform solid sphere? Should we get the same results if the sphere is a uniform spherical shell?

Thanks.

If this gives too much away please delete.
 
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  • #3
Should we get the same results if the sphere is a uniform spherical shell?
Yes. It is hard to provide the condition of non slipping in experiment

Little bit better: ball
simulation
 
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  • #4
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Yes. It is hard to provide the condition of non slipping in experiment

Little bit better: ball
simulation
Same setup as above but not as "perfect" orbits,



Interesting if correct simulation,
This is interesting and beyond me but not Google. I wonder if there are solutions to this problem where the ball orbits at constant distance from the tables rotation center? Thank you for the puzzle.

 
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Problem 3:

First of all, let me use a system of units where ##c=1##. Then the four-velocity is defined as
$$u^\mu = \frac{d x^\mu}{d\tau} = \gamma_u\begin{pmatrix}1\\\vec{u}\end{pmatrix}$$
where ##\vec{u}## is the regular 3-velocity and $$\gamma_u \equiv \frac{1}{\sqrt{1-\vec{u}\cdot\vec{u}}}$$
so, for a particle with 4-velocity ##u^\mu## in a reference frame ##\mathcal{S}##, the ##\gamma## factor is simply
$$\gamma_u = u^0$$.
Now, the Lorentz transformation to change from a RF ##\mathcal{S}## to another RF ##\mathcal{S}'##, where ##\mathcal{S}## moves with velocity ##\vec{v}=\begin{pmatrix}v\\0\\0\end{pmatrix}## with respect ##\mathcal{S}'## is given by
$$\Lambda(\mathcal{S}'\leftarrow \mathcal{S})^{\mu}{}_{\nu} =
\begin{pmatrix}\gamma_v & v\gamma_v&0&0 \\ v\gamma_v & \gamma_v&0&0\\
0&0&1&0\\
0&0&0&1\\\end{pmatrix}$$

So, because ##u^\mu## is a contravariant tensor
$$\gamma_{u'}=u'^0=\Lambda(\mathcal{S}'\leftarrow \mathcal{S})^{0}{}_{\mu}u^\mu = \gamma_v u^0 + v\gamma_v u^1 = \gamma_v\gamma_u + v\gamma_v \gamma_u u_x = \gamma_v\gamma_u\left(1+vu_x\right)$$
Recovering the ##c## we are left with
$$\gamma_{u'} = \gamma_v\gamma_u\left(1+\frac{vu_x}{c^2}\right)$$

I'm not sure about b) and c), and if they are right, probably there is a better way to do it, but here is my attempt:
The 4-position of the two clocks, in the RF of the left one (which I will call clock 1, while the other will be clock 2) are:
$$x_1^\mu(\tau)=\begin{pmatrix}\tau\\0\\0\\0\end{pmatrix}, \qquad x_2^\mu(\tau')=\begin{pmatrix}\tau'\\L\\y\\0\end{pmatrix}$$
Where ##\tau, \tau'## are the parametrizations of the worldlines. (I will keep an arbitrary ##y## by now.)
Then, in the rig reference frame (##\mathcal{S}##), which is related to the previous one by a ##u##-boost
$$\Lambda^{\mu}{}_{\nu} =
\begin{pmatrix}\gamma_u & 0&u\gamma_u&0 \\ 0 & 1&0&0\\
u\gamma_u&0&\gamma_u&0\\
0&0&0&1\\\end{pmatrix}$$
the 4-positions are
$$x_{1,\mathcal{S}}^\mu(\tau)=\gamma_u\tau\begin{pmatrix}1\\0\\u\\0\end{pmatrix}, \qquad x_{2,\mathcal{S}}^\mu(\tau')=\begin{pmatrix}\gamma_u(\tau'+ uy)\\L\\\gamma_u(u\tau'+ y)\\0\end{pmatrix}$$
And, in the last RF (##\mathcal{S}'##), related to ##\mathcal{S}## by the LT in part a) the 4-positions are
$$x_{1,\mathcal{S}'}^\mu(\tau)=\gamma_u\tau\begin{pmatrix}\gamma_v\\v\gamma_v\\u\\0\end{pmatrix}, \qquad x_{2,\mathcal{S}'}^\mu(\tau')=\begin{pmatrix}\gamma_v(\gamma_u\tau'+ u\gamma_uy + vL)\\\gamma_v(v\gamma_u\tau'+ vu\gamma_u y + L)\\\gamma_u(u\tau'+ y)\\0\end{pmatrix}$$
The amount of desyncronization in the RF ##\mathcal{S}'## will be the difference ##\Delta \tau = \tau'-\tau## in a fixed time (so, under the condition ##\Delta t=0##)
$$0=\Delta t = \gamma_v(\gamma_u\tau'+ u\gamma_uy + vL)-\gamma_u\tau\gamma_v \Longrightarrow \Delta \tau = -\frac{u\gamma_u y + vL}{\gamma_u}$$
with the clock 2 advanced an amout ##\Delta \tau##.

Now, b) and c) are the same problem with different initial conditions ##y##. The initial condition for b) is easy, simply we impose ##y=0##, and the desynchronization is given by the previous expression
$$\Delta \tau = -\frac{vL}{\gamma_u} \rightarrow -\frac{vL}{\gamma_u c^2}$$

For c) we need to impose that the two clocks are horizontally aligned in ##\mathcal{S}'## which means that
$$0=\Delta y = \gamma_u(u\tau'+ y)-u\gamma_u\tau = u\gamma_u \Delta \tau + y \gamma_u \Longrightarrow yu\gamma_u = -u^2\gamma_u \Delta \tau$$
And therefore
$$\gamma_u\Delta \tau =u^2\gamma_u \Delta \tau -vL\Longrightarrow \Delta \tau = -\gamma_u vL\rightarrow -\gamma_u \frac{vL}{c^2}$$

In both cases, ##v>0\Longrightarrow \Delta \tau <0## so clock 2 will be delayed, while ##v<0\Longrightarrow \Delta \tau > 0## and clock 2 will be advanced.
 
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  • #6
PeroK
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An attempt at Problem 2
Due to geometry of masses we would center of mass at the Center of the square frame. The line of action of force and position vector are parallel therefore there will no torque (moment).
The force ##F## will cause a translation of COM, the whole frame will move and by Newton’s Law the acceleration felt by any particle on the body is $$ a = \frac{F}{4m}##. Hence, do the acceleration of point ##C##.
 
  • #8
mmmmmmmmmm should I comment this? :)
 
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  • #9
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mmmmmmmmmm should I comment this? :)
Yes, please do comment.
 
  • #10
Spinnor
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The following is an approximate realization of problem 2. Please delete is it gives away too much.

 
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  • #11
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The following is an approximate realization of problem 2. Please delete is it gives away too much.

Did you fix your camera on your fan?

By the way thank you so much for the demonstration, I didn’t take into account that deformation of the frame was allowed. As far as I can think the point opposite to where the force is applied must accelerate away from the direction of force (due to formation).
 
  • #12
Spinnor
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Did you fix your camera on your fan?

By the way thank you so much for the demonstration, I didn’t take into account that deformation of the frame was allowed. As far as I can think the point opposite to where the force is applied must accelerate away from the direction of force (due to formation).
That was my first thought, the fan, but I found a microphone stand which you can make out a bit.

I think from the experiment there may be 3 different cases depending on the initial shape, squished one way, right angles, squished the other way?
 
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  • #13
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We can break down the force ##\vec{F}## into components, one along AB and the other along AD.
image.jpg

Since, the diagonal of a square makes an angle of ##45^{\circ}## with its sides, therefore by geometry we would get $$ F_1 = F_2 = \frac{F}{\sqrt 2}$$.

SCENARIO 1
Force ##F_2## will cause AB to rotate anti-clockwise about its COM (which lies at its center). Moment produced by ##F_2## is ##\frac{b~F}{2\sqrt 2}##. Let the angular momentum of the rod be denoted by ##L##, then
$$ \frac{dL}{dt} = \frac{b~F}{2\sqrt 2} \\
\alpha = \frac{b~F}{2\sqrt 2 m} $$

The acceleration that point B will experience along BC (relative to COM of AB) is $$ a_{relative} = \alpha ~ b/2 \\ a_{relative} = \frac{b^2 ~F}{4\sqrt 2 m}$$
Acceleration of COM in a direction directly opposite to BC is ##\frac{F}{\sqrt 2 m}##. So, actual acceleration of point B is $$ a_B = \frac{b^2 ~F}{4\sqrt 2 m} - \frac{F}{\sqrt 2 m} = \frac{F}{4\sqrt 2} (b^2-4)$$.

So, acceleration on point C along the direction of BC (B to C) is ##a_B##. Force ##F_1## acts parallel to AB and will cause the rotation of the arm BC. Moment produced by ##F_1## on the arm BC is ##\frac{b~F}{2\sqrt 2}##. Angular acceleration of BC $$ \frac{dL}{dt} = \frac{b~F}{2\sqrt 2} \\
\alpha = \frac{b~F}{2\sqrt 2 m} $$
Acceleration of point C (relative to COM of BC) along the direction of DC (D to C) is $$ a_{relative}= \frac{b^2 ~F}{4 \sqrt 2 m} $$ and the actual acceleration of point C is $$ a_c = \frac{b^2~F}{4\sqrt 2 m} - \frac{ F}{\sqrt 2 m} \\
a_c = \frac{F}{4\sqrt 2} (b^2 -4)$$
Free body diagram of arm BC:
image.jpg
Since, ##a_c= a_B##, therefore the result acceleration will make an angle of 45 degrees with both ##a_c## and ##a_B## and hence the net acceleration of point C is along the diagonal AC (A to C) and it’s magnitude is ## a_{net}=\frac{F}{4m} (b^2-4)##.

SCENARIO 2
Due to symmetry, the rotation of arms AD and CD will produce the same net acceleration on C and in the same direction.

FINAL ANSWER
So, the acceleration of the point C right after the force was applied is $$ a = \frac{F}{2m} (b^2-4) $$
 
  • #14
etotheipi
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Though doesn't the wording a force ##F## acts on the hinge imply a force of only ##\frac{F}{2}## on AB and AD each here (in addition to a vertical reaction force due to the hinge itself). And then also reaction forces due to the hinges at the other end?
 
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  • #15
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b is a length, right?
 
  • #16
Solution of the problems implies the following two stages:

1) to write down differential equations of motion based on the laws and theorems of dynamics;

2) to provide a mathematically correct deduction of a result from the equations of motion.
 
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  • #17
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Are we encouraged to point out issues, or would you prefer just solutions be submitted?
 
  • #18
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Is my solution of problem 2 in post #13 incorrect ?
 
  • #19
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Is my solution of problem 2 in post #13 incorrect ?
The equation for α is dimensionally inconsistent as is the result.
 
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  • #20
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Error has been fixed, in post #13 I mistakingly took inertial mass instead of Moment of Inertia during the calculation of angular momentum.
We can break down the force ##\vec{F}## into components, one along AB and the other along AD.
View attachment 261267

Since, the diagonal of a square makes an angle of ##45^{\circ}## with its sides, therefore by geometry we would get $$ F_1 = F_2 = \frac{F}{\sqrt 2}$$.

SCENARIO 1
Force ##F_2## will cause AB to rotate anti-clockwise about its COM (which lies at its center). Moment produced by ##F_2## is ##\frac{b~F}{2\sqrt 2}##. Let the angular momentum of the rod be denoted by ##L##, then
$$ \frac{dL}{dt} = \frac{b~F}{2\sqrt 2} \\
\alpha = \frac{6F}{\sqrt 2 mb}$$

The acceleration that point B will experience along BC (relative to COM of AB) is $$ a_{relative} = \alpha ~ b/2 \\ a_{relative} = \frac{3F}{\sqrt 2 m}$$
Acceleration of COM in a direction directly opposite to BC is ##\frac{F}{\sqrt 2 m}##. So, actual acceleration of point B is $$ a_B = \frac{3F}{\sqrt 2 m} - \frac{F}{\sqrt 2 m} = \frac{\sqrt 2}{m}$$.

So, acceleration on point C along the direction of BC (B to C) is ##a_B##. Force ##F_1## acts parallel to AB and will cause the rotation of the arm BC. Moment produced by ##F_1## on the arm BC is ##\frac{b~F}{2\sqrt 2}##. Angular acceleration of BC $$ \frac{dL}{dt} = \frac{b~F}{2\sqrt 2} \\
\alpha = \frac{6F}{\sqrt 2 mb}$$
Acceleration of point C (relative to COM of BC) along the direction of DC (D to C) is $$ a_{relative}= \frac{3F}{\sqrt 2 m} $$ and the actual acceleration of point C is $$ a_c = \frac{3F}{\sqrt 2 m}- \frac{ F}{\sqrt 2 m} \\
a_c = \frac{\sqrt 2 F}{m}$$
Free body diagram of arm BC:
View attachment 261273
Since, ##a_c= a_B##, therefore the result acceleration will make an angle of 45 degrees with both ##a_c## and ##a_B## and hence the net acceleration of point C is along the diagonal AC (A to C) and it’s magnitude is ## a_{net}=\frac{2F}{m}##.

SCENARIO 2
Due to symmetry, the rotation of arms AD and CD will produce the same net acceleration on C and in the same direction.

FINAL ANSWER
So, the acceleration of the point C right after the force was applied is $$ a = \frac{4F}{m}$$
 
  • #22
Spinnor
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Thus I see that there is no interest in this thread
There is some interest, I pulled out my Engineering Dynamics book to study the relevant physics and at first could not understand the simplest simplification of this problem, a single rod with one end constrained to move on a line with a force applied to it, I think I get it now. In the process I think I know a little more and will give the 2nd problem one more try today (I promise I won't look at the solution.)
 
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  • #23
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Thus I see that there is no interest in this thread, so to finish with that the solution of the second problem is contained here https://www.physicsforums.com/threads/aceeleration-of-a-vertex.672835/#post-6331831
Problem 2 has taught me so many things that I think I could never learn from any book. The only problem was that I didn’t know Lagrangian Mechanics, although the problem could be solved by Newtonian Mechanics but solving it using NM wasn’t a good way.

I vote for physics challenge to be continued.
 
  • #24
etotheipi
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I think it's really hard to get the balance right with these sorts of challenges. I'm not quite too sure where the "Goldilocks" zone is!

Like Adesh said it'd be great to have something like this going, but it's a lot of effort on the problem setters' part and you'd want to make sure the thread was getting good traction. Either way, thanks to @wrobel and @PeroK for these questions!
 
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  • #25
TSny
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PROBLEM 1

1587935951032.png


Let ##\vec R_S## be the position of the center of the ball relative to the origin ##O##.

Let ##\vec r## be the position vector of ##P## relative to the center of the ball ##S##. ##\vec r## always points vertically downward and has magnitude ##r ## equal to the radius of the ball. So, ##\vec r## does not depend on time.

If ##\vec \omega## is the angular velocity of the ball about its center, then ##S## has velocity ##\vec r \times \vec \omega## relative to the rotating plane, assuming rolling without slipping. The rotation of the plane adds an additional amount to the velocity of ##S## relative to the lab. This additional amount is given by ##\vec \Omega \times \vec R_s##. So, the velocity of ##S## relative to the lab is $$\vec V_S = \vec r \times \vec\omega + \vec \Omega \times \vec R_S$$

The acceleration of ##S## relative to the lab is

$$\vec a_S = \frac{d \vec V_S}{dt} = \vec r \times \frac{d \vec \omega}{dt} + \vec \Omega \times \frac{d \vec R_S}{dt}$$

$$\vec a_S = \vec r \times \vec \alpha + \vec \Omega \times \vec V_S $$

##\vec \alpha## is the angular acceleration of the ball about its center ##S## and is due to the torque ##\vec \tau = \vec r \times \vec f##, where ##\vec f## is the force of friction acting on the ball. If ##I_S ## is the moment of inertia of the sphere about its center, ##\vec \alpha = \frac{\vec r \times \vec f}{I_S}##.

So, $$\vec a_S = \frac { \vec r \times \left( \vec r \times \vec f \right)} {I_S} + \vec \Omega \times \vec V_S $$

Newton’s second law in the lab frame gives ##\vec f = m \vec a_S##, where ##m## is the mass of the ball. Thus, $$\vec a_S = \frac {m}{I_S} \vec r \times \left( \vec r \times \vec a_S \right) + \vec \Omega \times \vec V_S $$

Note that ##\vec r \times \left( \vec r \times \vec a_S \right) = -r^2 \, \vec a_S## since ##\vec a_S## and ##\vec r## are perpendicular to each other. Hence, $$\vec a_S = -\frac {m r^2}{I_S} \vec a_S + \vec \Omega \times \vec V_S = -\frac{5}{2} \vec a_S + \vec \Omega \times \vec V_S $$ Here, we used ##\frac {m r^2}{I_S} = \frac{5}{2}## for a solid sphere.

Solving for ##\vec a_S##, $$\vec a_S = \frac{2}{7} \vec \Omega \times \vec V_S $$

Thus, the acceleration of ##S## relative to the lab is always perpendicular to the velocity of ##S## relative to the lab. Since ##\vec \Omega## is constant, we may conclude that S moves in uniform circular motion relative to the lab. Letting ##\rho## denote the radius of the circular motion, ##a_S= \frac{V_S^2}{\rho}##.

So, $$ \frac{V_S^2}{\rho} = \frac{2}{7} \Omega V_S $$

and, $$\rho = \frac{7}{2} \frac{V_s}{\Omega}$$.

The radius of the circular motion depends only on ##\Omega## and the speed ##V_S## of the ball relative to the lab.

Aside: The orbital angular speed of the circular motion is ## \omega_{\rm orb} = V_s/\rho = \frac{2}{7} \Omega##. This appears to agree fairly well with what you see in the last video of post #4.

For a hollow sphere with ##I_S = \frac{2}{3}mr^2##, one obtains $$\rho = \frac{5}{2} \frac{ V_s}{ \Omega}$$ So, in this case, ## \omega_{\rm orb} = \frac{2}{5} \Omega## , which agrees pretty well with the data at the end of the second video in post # 4 where a tennis ball was used.
 
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