Rolling stone

http://faculty.missouristate.edu/l/lesreid/potw.html

Missouri State University’s Advanced Problem # 104

Vincent Pantaloni (Orléans, France)

Problem : A circle one foot in radius is placed on a ramp that makes an angle of 30 degrees with the horizontal so that a point P on the circle touches the ramp 10 feet from the base of the ramp. As the circle rolls down the ramp, what is the maximum height that P attains ?

10

1

30°

P

Solution :

Here’s a picture of what is happening. We know that point P will draw an arch of cycloid (dotted curve).

It’s highest point will be reached when we have a horizontal tangent to that arch. We need to find that point P^′ and then compute the length BP^′, that is BA+AP^′. I don’t want to find an equation to that arch on a slope, I’d rather turn my head 30 degrees

to the left and look at it this way : _30°

P Horizontal tangent to the arch

P'

B A

O

After this 30° rotation our "horizontal tangent"

has a slope oftan(−30) =−^√13. In this situation, Wikipedia reminds me that such a cycloid arch has parametric equation :

x=t−sint

y= 1−cost , t∈[0,2π]

1 2

1 2 3 4 5 6

b

b= 4.44

b

P^′

b

A

b

P We are now searching the value oftfor which we have a tangent vector whose coordinates are proportional to (1,−^√13). The tangent vector has coordinates :

x^′ = 1−cost

y^′ = sint , t∈[0,2π]

These coordinates are proportional iff the cross products are equal, so we have to solve : sint=− 1

√3(1−cost) (1) Hopefuly we’ll make it look better. . . All of the following equations are equivalent :

(1) ⇐⇒ cost−1−√

3 sint= 0 (2)

⇐⇒ 1 2cost−

√3

2 sint= 1

2 (3)

⇐⇒ cos t+π

3 =1

2 (4)

⇐⇒ t+π 3 =±π

3 [2π] (5)

(6)

http://prof.pantaloni.free.fr mail me

http://faculty.missouristate.edu/l/lesreid/potw.html

The only solution witht∈]0,2π[is−^2π3 + 2π= ^4π3 . By substituting in the cycloid’s parametric equation, we get the coordinates of P^′ on the graph : P^′(^4π₃ +^√₂³,³₂)which is approximately(5.05,1.5) and that seems quite consistent with our graph. Now in order to findAP^′we have to search for the coordinates of pointA. Now it’s quite easy, just a matter of straight lines :

The line(AP^′)is perpendicular to our tangent, so it has a slope of√

3.(AP^′) :y=√

3x+b. We find the y intersect by using the fact thatP^′ is on this line so :

yP′ =√

3xP′+b that is 3 2 =√

3 4π 3 +

√3 2

! +b

Solving forbgivesb=−4π√ 3

3 . NowAhas coordinatesA(xA,0), we look forxA : yA=√

3xA+b gives xA=− b

√3 =4π 3 We can now computeAP^′ :

AP^′ = q

(xP′−xA)²+ (yP′ −yA)²= v u u t

√3 2

!² +

3 2

²

=√ 3

Now we needABso first we haveAP =xP−xA= 2π−^4π3 =^2π3 and then knowing thatpoint P on the circle touches the ramp 10 feet from the base of the rampwe haveOA=OP−AP = 10−^2π3. Considering the right angle triangleOAB we have :

AB=OAsin(30) =

10−2π 3

×1

2 = 5−π 3

And finally adding AB andAP^′ we find BP^′ which is what we were looking for. The maximum height that P attains is :

BP^′=BA+AP^′ = 5−π 3 +√

3 That is just√

3−^π3 ≃0.68ft above it’s starting point.

Remarks. The result seems quite simple compared to the computations I had to do, so there might be a simpler geometrical solution. One might think thatP^′is the first position where the point P is "on top"

of the circle. Actually it is not (if my computations are correct) because when this happens the length AP is equal toπ−^π6 = ^5π6 which is not the same as what we found : AP = ^2π3 . You can see that this first guess is wrong on the picture below (done using GeoGebra) :

b

b

b b

b b

b

http://prof.pantaloni.free.fr mail me