A roller coaster moving along its track rolls into a circular loop of radius r. In the loop, it is only affected by its initial velocity, gravity, and the shape of the track. Let v denote the instantaneous speed and a denote the magnitude of the instantaneous acceleration of the roller coaster in the loop. Which of the following is true in the loop?
a. The roller coaster is not in uniform circular motion, but we still have a=v^2/r everywhere on the loop
b. The roller coaster is not in uniform circular motion, but the tangential acceleration is so small that we can approximate a by v^2/r everywhere on the loop
c. The roller coaster is in uniform circular motion
d. The roller coaster is not in uniform circular motion, and a=v^2/r is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes

Since the angular speed, ω = v/r does not change, the magnitude of its tangential acceleration is zero although there is a change in its direction because the direction of its initial velocity changes. That is a" = rα and α = Δω/Δt since Δω = 0, α = 0 and a" = r(0) = 0

So, there is no tangential acceleration. Since there is no tangential acceleration, our instantaneous acceleration which is the vector sum of our centripetal acceleration and tangential acceleration is a = √(a'² + a"²) = √(a'² + 0²) = √a'² = a' = v²/r

So, a is always v²/r.

Since the instantaneous acceleration is always (a = v²/r) constant, the motion is uniform. So, it is uniform circular motion.

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-10-26T17:18:03+02:00

The roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.

The given parameters;

radius of the circular path, = r
instantaneous speed = v
instantaneous acceleration = a

The motion tension on the loop at the lowest point in the circular motion is given as;

[tex]T = mg + \frac{mv^2}{r}[/tex]

The motion tension on the loop at the highest point in the circular motion is given as;

[tex]T = \frac{mv^2}{r} - mg[/tex]

This shows that circular motion is affected by;

acceleration due to gravity, g
radius of the circular path, r
speed of the motion, v
mass of the object, m

Thus, we can that the roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.

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