Question

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.

Answer 1

h >5/2r

Step-by-step explanation:

This problem involves the application of the concepts of force and the work-energy theorem.

The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.

Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)

So from newton's second law of motion,

W – N = mv²/r

N = normal force = 0

W = mg

mg = ma = mv²/r

mg = mv²/r

v²= rg

v = √(rg)

The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop

So

ΔPE = ΔKE = 1/2mv²

The height at the roller coaster starts is usually higher than the top of the loop by design. So

ΔPE =mgh - mg×2r = mg(h – 2r)

2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.

In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.

So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.

ΔPE > ΔKE

mg(h–2r) > 1/2mv²

g(h–2r) > 1/2(√(rg))²

g(h–2r) > 1/2×rg

h–2r > 1/2×r

h > 2r + 1/2r

h > 5/2r

Answer 2

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.

Step-by-step 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.