Quiz: Wheels

[axpgn]

A car is being driven so that its wheels, all of radius [size=150]$a$[/size] feet, have an angular velocity of [size=150]$omega$[/size] radians per second.
A particle is thrown off from the tire of one of these wheels, where it is supposed that [size=150]$aomega^2>g$[/size].
Neglecting the resistance of the air, show that the maximum height above the roadway which the particle can reach is [size=150]$(aomega+gomega^(-1))^2/(2g)$[/size]



Cordialmente, Alex

[ingres]

La particella dovrà trovarsi sulla parte superiore della ruota in una posizione angolare $theta$ rispetto al centro della ruota.
Quindi la sua altezza iniziale sarà $a+a*sin(theta)$ e la sua velocità in senso verticale $omega*a*cos(theta)$
Tale velocità gli consentirà di raggiungere un'ulteriore quota rispetto al punto iniziale di

$Delta h= (omega*a*cos(theta))^2/(2g)$ per la conservazione dell'energia.

Quindi l'altezza totale sarà:

$h = a+a*sin(theta)+(omega*a*cos(theta))^2/(2g)$

derivando rispetto a $theta$ il valore massimo si ha per

$sin(theta) = g/(omega^2*a)$ (tale soluzione esiste per l'ipotesi che $g
valore che sostituito nell'espressione di h consente di trovare la formula data.

[moccidentale]

.

[axpgn]

Siccome mi ci vuole tempo per capire ( ) vi posto la soluzione, per così dire, "ufficiale" così vedete le differenze (se ci sono)
Proviene da una gara degli anni trenta.

Solution:
If a particle is thrown into motion in a gravitational field starting at height $h$ and with upward component of velocity $v$, it will rise to the height $h + (v^2/(2g))$ [The horizontal components of the motion do not in-
fluence the maximum height].

As long as the particle remains attached to the tire, it follows the path of a cycloid, and we may take its equations of motion as
$x = aomegat - a sin(omegat)$
$y = a(1 - cos(omegat))$

If the particle leaves the tire when $omegat = theta$, then it starts into gravitational motion with height $a(1-cos(theta))$ and upward velocity component $y'=aomegasin(theta)$.
Hence it reaches the height $H=a(1-cos(theta))+(a^2omega^2)/(2g)sin^2theta$ [N1]
provided $0 < theta < pi$ (to ensure that the particle starts upward).

We are asked to maximize $H$ by choice of $theta$.

We set $(dH)/(d theta)=asin(theta)+(a^2omega^2)/gsin(theta)cos(theta)=0$
and find that the critical points are $0, pi, theta_0$ , where $theta_0 = arccos (-g/(aomega^2))$.
(Since $aomega^2>g$, there is such a $theta_0$.)

The corresponding values of $H$ are $0, 2a$ and $H_0=a(1+g/(aomega^2))+(a^2omega^2)/(2g)(1-g^2/(a^2omega^4))=1/(2g)(aomega+gomega^(-1))^2$

Since $(aomega + gomega^(-1))^2 > 4ag, H_0 > 2a$, so the maximum value of $H$ is $H_0$.


We can also find the maximum value of $H$ by writing [N1] in the form $H=H_0-(a^2omega^2)/(2g)(cos(theta)+g/(aomega^2))^2$


REMARK.
If $aomega^2 <= g$, the maximum value of $H$ will be 2a, attained for $theta=pi$, which means that particles flying off never go higher than the top of the wheel.

Cordialmente, Alex

[moccidentale]

.

[axpgn]

Carina