Question

As a roller coaster car crosses the top of a 48.01-m-diameter loop-the-loop, its apparent weight is the same as its true weight. What is the car's speed at the top?

Answer 1

Answer:

The speed of the car, v = 21.69 m/s

Explanation:

The diameter is  = 48.01 m

Therefore, the radius of the loop R = 24.005 m

Weight at the top is n = mv^2/R - mg

Since the apparent weight is equal to the real weight.

So, mv^2/R - mg = mg

v = √(2Rg)

v = √[2(24.005 m)(9.8 m/s^2)]

The speed of the car, v = 21.69 m/s

Answer 2

Answer:

The speed is 15.34 m/s.

Explanation:

Diameter, d = 48.01 m

Radius, R = 24.005 m

Let the speed is v and the mass is m.

Here, the weight of the car is balanced by the centripetal force.

According to the question

$m g = \frac{mv^2}{R}\\\\v =\sqrt{24.005\times9.8}\\\\v = 15.34 m/s$