You are an engineer helping to design a roller coaster that carries passengers down a steep track and around a vertical loop. The coaster's speed must be great enough when at the top of the loop so that the rider stays in contact with the cart and the cart stays in contact with the track. Riders can withstand acceleration no more than a few "gg's", where one "gg" is 9.89.8 m/s2.What are some reasonable values for the physical quantities you can use in the design of the ride? For example, one consideration is the height at which the cart and rider should start so that they can safely make it around the loop and the radius of the loop.

Maximum height was designed to be 20m while minimum height was designed as 12m. Thus, the height should not be less than 12m otherwise, the cart will travel too slowly around the loop and will not remain in contact with the track.

Explanation:

I've attached a force diagram of the roller coaster layout at crest.

In the diagram ;

F_SonP = The force exerted by the seat on the person in the cart.

F_EonP = Force exerted by earth on person

Now, if we apply the component form of Newton's 2nd law of motion the vertical direction at crest position, and equate the net applied force to the centripetal force term due to circular motion about a radius(r), we obtain;

ΣF_onC = ma

Centripetal accelaration = v²/r

Thus,

So, N_SonP - F_EonCy = - m(v_f²/r)

And so,

N_SonP - mg = - m(v_f²/r)

N_SonP = m(g - (v_f²/r) - - - - - (eq1)

At minimum limiting velocity,

N_SonP < 0

So, m(g - (v_f²/r) < 0

Making v_f the subject;

g < (v_f²/r)

v_f² > rg

v_f >√rg - - - - (eq2)

Also, at upper maximum velocity, let's say the person's maximum limit of acceleration has a value of 3g.

Thus,

v_f < √3gr - - - - (eq3)

Thus, we can write the range limits as;

√rg < v_f < √3gr - - - - (eq4)

Now, if we assume that the friction has a negligible friction, we can apply the work energy principle to obtain;

U_i + W = U_f

This gives;

K_i + U_gi + W = K_f + U_gf

K_i, W and U_gf are zero. Thus,

U_gi = K_f

Now, due to the minimum and maximum limits, U_gi = ΔU_g =

-mg(Δh)

Thus,

mg(Δh) = K_f

mg(Δh) = (1/2)mv_f²

g(Δh) = (1/2)v_f²

Now, (Δh) = r - h

Thus, g(r-h) = (1/2)v_f² - - - - eq(5)

Where h is the height at which the cart and rider should start.

r is radius loop. Let's assume it has a value of 8 m. Thus, if we plug it into eq(4),we obtain;

√8g < v_f < √24g

g = 9.8 m/s²

√8 x 9.8 < v_f < √24 x 9.8

8.85 m/s < v_f < 15.34m/s

Now, since the maximum velocity is 15.34 m/s,lets plug it in eq(5) to obtain;

-9.8(8-h) = (1/2)•15.34²

-9.8(8-h) = 117.6578

8-h = -117.6578/9.8

8 + 12 = h

h = 20m

Now, since the minimum velocity is 8.85 m/s,lets plug it in eq(5) to obtain;

-9.8(8-h) = (1/2)•8.85²

-9.8(8-h) = 39.16125

h - 8 = 39.16125/9.8

h = 8 + 4 = 12m

So height should not be less than 12m otherwise, the cart will travel too slowly around the loop and will not remain in contact with the track.

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