Just a cool high school problem

[wrobel]

Thought some teachers might find this useful.

A wheel of radius $R$ rotates in a vertical plane about its fixed center. A speck of dirt detaches from a point on the wheel and, after a time interval $T$, impacts the wheel at the same height. Determine the angular velocity of the wheel. The acceleration due to gravity is $g$.
Find all possible answers.

 

[berkeman]

Find all possible answers.

Spoiler

I can see a series of solutions, but they all have the speck leaving the rim flying up vertically when the spot on the wheel is at 90 degrees from the vertical and the spot is moving up. The series of solutions are for one wheel rotation until re-impact, two rotations, and so on. Are there solutions other than from that starting position?

 

[wrobel]

Only two solutions: the point moves strictly vertically up and down, or it traces a parabola.

 

[Halc]

It seems that any angular velocity will do so long as it's high enough to involve centripetal acceleration of at least g. The problem needs perhaps another bit of information to narrow this down.

 

[DaveE]

Only two solutions: the point moves strictly vertically up and down, or it traces a parabola.

Yes, so there is no unique answer for the purely vertical motion of the particle. It will hit the wheel at the same height it left without regard to the velocity. I suggest removing this trivial solution.

Anyway, just to clarify. "at the same height" has nothing to do with the same location on the wheel, it's just relative to an external observer? You didn't mean at the same location on the wheel, right?

 

[wrobel]

Yes, so there is no unique answer for the purely vertical motion of the particle.

There is a unique answer for the purely vertical motion: $\omega=\frac{gT}{2R}$
$$-\frac{gT^2}{2}+\omega R T=0$$
Just read the text carefully

 

[willyengland]

I don't understand the question.
The dirt particle sits on the rim of the wheel, is flung upwards, and flies off. Then it falls vertically back down and lands on the wheel.
Is that the idea?

 

[Roberto Pavani]

Nice problem.

The clean way to solve the problem is to use an inertial reference frame whose origin is the center of the wheel. This frame is not rotating; it only translates with the center of the wheel climbing a wall, therefore the acceleration of gravity is still exactly g.

Let the point where the particle detaches have height h above the center of the wheel, with:

$0<h<R$

We choose the detachment point on the left side of the wheel. The initial coordinates are:

$x_0=-\sqrt{R^2-h^2}$

(the wall is on the right side, with vertical velocity -V in this reference description, and the wheel rotates clockwise).

The translational velocity of the wheel center is:

$V=\omega R$

The velocity of a point on the circumference of the wheel is the sum of the translational velocity of the center and the rotational velocity around the center:

$\vec v_{\rm rim}=\vec V+\vec\omega\times\vec r$

When we use the inertial frame attached to the wheel center, the translational velocity is removed. The particle leaves the wheel with the tangential velocity due to the rotation.

Therefore the initial velocity components are:

$v_x=\omega h$

and:

$v_y=\omega\sqrt{R^2-h^2}$


The vertical motion must take into account that the wheel center moves during the flight.

In the reference frame of the wheel center, the particle starts at height h with vertical velocity:

$v_y=\omega\sqrt{R^2-h^2}$

The vertical motion is:

$y(t)=h+v_y t-\frac{1}{2}gt^2$


However, after a time T the wheel center has moved vertically by:

$VT$

Therefore, although the particle returns to the same absolute height in the ground frame, its final height relative to the moving wheel center is:

$y(T)=h-VT$


The flight time is obtained from:

$h+v_yT-\frac{1}{2}gT^2=h-VT$


which gives:

$(v_y+V)T-\frac{1}{2}gT^2=0$


The non-zero solution is:

$T=\frac{2(v_y+V)}{g}$


Using:

$v_y=\omega\sqrt{R^2-h^2}$

and:

$V=\omega R$

we obtain:

$T=\frac{2\omega(R+\sqrt{R^2-h^2})}{g}$


The horizontal coordinate after this time is:

$x(T)=x_0+v_xT$


therefore:

$x(T)=-\sqrt{R^2-h^2}+\omega hT$


For the particle to hit the wheel again, the final point must belong to the circumference of the wheel.

The final coordinates relative to the wheel center are:

$(x(T),h-VT)$


Therefore the geometrical impact condition is:

$x(T)^2+(h-VT)^2=R^2$


Substituting the expressions for x(T) and V:

$\left(-\sqrt{R^2-h^2}+\omega hT\right)^2+(h-\omega RT)^2=R^2$


Together with:

$T=\frac{2\omega(R+\sqrt{R^2-h^2})}{g}$
or
$\omega =\frac{g T} {2 (R+\sqrt{R^2-h^2})}$

this equation determines the possible angular velocities.


The important point is that the translation of the wheel cannot be ignored. The particle returns to the same absolute height in the ground frame, but because the wheel center has moved, its height relative to the moving wheel is:

$h-VT$

which is the quantity that must be used in the final intersection condition.

 

[wrobel]

The translational velocity of the wheel center is

fixed center

Honestly, I'm a bit shocked by everything that's going on here.

$$x(t)=-R\omega t\sin\varphi+R\cos \varphi$$
$$y(t)=-gt^2/2+R\omega t\cos\varphi+R\sin\varphi$$
$$x(T)=-x(0),\quad y(T)=y(0)$$
Two equations, two unknowns: $\varphi,\omega$

 

[wrobel]

edited
edited

 

[Roberto Pavani]

Correction: I had initially visualized a wheel translating upward while rotating. Re-reading the statement, the center is fixed and the wheel only rotates.

Therefore:

$h + v_y T - \frac{1}{2} g T^2 = h$​

which gives:

$v_y = \frac{gT}{2}$​

For a detachment point at height h:

$v_y = \omega \sqrt{R^2-h^2}$​

hence:

$\omega \sqrt{R^2-h^2} = \frac{gT}{2}$​

Since the particle starts at:

$x_0=-\sqrt{R^2-h^2}$​

and must hit the wheel again at the same height:

$x_f=+\sqrt{R^2-h^2}$​

the horizontal displacement condition is:

$\omega h T = 2\sqrt{R^2-h^2}$​

Combining the two equations:

$\frac{gT}{2\sqrt{R^2-h^2}} = \frac{2\sqrt{R^2-h^2}}{hT}$​

which leads to:

$4h^2 + gT^2 h - 4R^2 = 0$​

and therefore:

$h = \frac{-gT^2 + \sqrt{g^2T^4+64R^2}}{8}$​

Finally:

$\omega = \frac{gT}{2\sqrt{R^2-h^2}}$​

or, eliminating h:

$\omega = \frac{4gT}{\sqrt{128R^2-g^2T^4+gT^2\sqrt{g^2T^4+64R^2}}}$​

 

[A.T.]

I can see a series of solutions, but they all have the speck leaving the rim flying up vertically when the spot on the wheel is at 90 degrees from the vertical and the spot is moving up.

Yes, so there is no unique answer for the purely vertical motion of the particle. It will hit the wheel at the same height it left without regard to the velocity. I suggest removing this trivial solution.

$T$ is fixed.

 

[JimWhoKnew]

Honestly, I'm a bit shocked by everything that's going on here.

OP doesn't say that the speck detaches from, and lands back on the rim of the wheel. I visualized the wheel as a full disc, which makes the problem more complex.

 

[willyengland]

OP doesn't say that the speck detaches from, and lands back on the rim of the wheel. I visualized the wheel as a full disc, which makes the problem more complex.

Agree; the question wasn't clear, which is why I asked for clarification. An illustration would have been helpful.

 

[gyorgiy]

Presumably ignore air resistance...

 

[Roberto Pavani]

Short explanation of my calculation in #11:
Ignoring the trivial vertical case, I considered a particle at height h on the wheel.
At detachment its speed is $V=\omega R$ and, from that instant on, its motion is purely ballistic.
The subsequent rotation of the wheel is irrelevant: an equivalent problem is to freeze the wheel at the instant of detachment.
Since the projectile must return to the circle at the same height h, the only possible landing point is the symmetric point on the opposite side:
$x_i=-\sqrt{R^2-h^2} \qquad x_f=+\sqrt{R^2-h^2}$
hence:
$\Delta x=2\sqrt{R^2-h^2}$
which is the starting point of the derivation given in #11.