Question

A roller coaster starts at the top of a hill of height h, goes down the hill, and does a circular loop of radius r before continuing on. This roller coaster is not attached to the track — it only glides on top of it, so it could fall off at the top of the loop. This section of the roller coaster is shown below.
If the roller coaster starts at the top of the hill with zero velocity, what is the expression for the velocity of the roller coaster at the top of the loop?
a. Draw the free-body diagram for the roller coaster at the top of the loop. (1 point)

b. Write the expression for the net force in the y-direction at the top of the loop, in terms of w, FN, m, v, and r. (1 point)

c. Write the equation that will tell you the velocity at which the coaster will fall when it reaches the top of the loop. (1 point)

d. If the coaster has a mass of 50 kg and the loop has a radius of 30 m, how fast must the coaster be going at the top of the loop to keep it from falling? (1 point)

e. Based on the values in the previous part, how high must the roller coaster start out to prevent it from falling from the top of the loop? (2 points)

f. If friction along the track caused the coaster to lose 10% of its initial energy by the time it reached the top of the loop, how high would h need to be to keep the coaster from falling off? (2 points)

Answer 1

a) See free-body diagram in attachment

b) Net force in the y-direction: $F_y=mg+N$[/tex]

c) The velocity at which the roller coaster will fall is $[tex]v=\sqrt{gr}$[/tex]

d) The speed of the roller coaster must be 17.1 m/s

e) The roller coaster should start from a height of 90 m

f) The roller coaster should start from a height of 100 m

Explanation:

a)

See the free-body diagram in attachment. There are only two forces acting on the roller coaster at the top of the loop:

• The weight of the roller coaster, acting downward, indicated by $mg$ (where m is the mass of the roller coaster and g is the acceleration of gravity)
• The normal reaction exerted by the track on the roller coaster, acting downward, and indicated with N

The two forces are represented in the diagram as two downward arrows (the length is not proportional to their magnitude, in this case)

b)

Since there are only two forces acting on the roller coaster at the top of the loop, and both forces are acting downward, then we can write the vertical net force as follows (we take downward as positive direction):

$F_y = mg + N$

where

mg is the weight

N is the normal reaction

Since the roller coaster is in circular motion, this net force must be equal to the centripetal force, therefore

$m\frac{v^2}{r}=mg+N$

where v is the speed of the car at the top of the loop and r is the radius of the loop.

c)

For this part of the problem, we start from the equation written in part b)

$m\frac{v^2}{r}=mg+N$

where the term on the left represents the centripetal force, and the terms on the right are the weight and the normal reaction.

We now re-arrange the equation making v (the speed) as the subject:

$v=\sqrt{gr+\frac{Nr}{m}}$

However, the velocity at which the roller coaster will fall is the velocity at which the normal reaction becomes zero (the roller coaster loses contact with the track), so when

N = 0

And as a result, the minimum velocity of the cart is

$v=\sqrt{gr}$

d)

In this part, we are told that the radius of the loop is

r = 30 m

And the mass of the cart is

m = 50 kg

Moreover, the acceleration of gravity is

$g=9.8 m/s^2$

We said that the minimum velocity that the cart must have in order not to fall at the top is

$v=\sqrt{gr}$

And substituting, we find

$v=\sqrt{(9.8)(30)}=17.1 m/s$

e)

According to the law of conservation of energy, the initial gravitational energy of the roller coaster at the starting point must be equal to the sum of the kinetic energy + gravitational potential energy at the top of the loop, therefore:

$mgh = \frac{1}{2}mv^2 + mg(2r)$

where

h is the initial height at the starting point

(2r) is the height of the roller coaster at the top of the loop

We can re-arrange the equation making h the subject,

$h=\frac{v^2}{2g}+2r$

And substituting the minimum speed of the cart,

$v=\sqrt{gr}$

this becomes

$h=r+2r=3r$

And since r = 30 m, we find

$h=3(30)=90 m$

f)

In this case, 10% of the initial energy is lost during the motion of the roller coaster. We can rewrite the equation of the previous part as

$0.90mgh = \frac{1}{2}mv^2 + mg(2r)$

Because only 90% (0.90) of the initial energy is converted into useful energy (kinetic+potential) when the cart reaches the top of the loop.

Re-arranging the equation, this time we get

$h=\frac{\frac{v^2}{2g}+2r}{0.90}$

Again, by substituting $v=\sqrt{gr}$, we get

$h=\frac{3r}{0.90}$

And therefore, the new initial height must be

$h=\frac{3(30)}{0.9}=100 m$

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