Question

A roller coaster is created with a hill of height 125 m and a loop with radius 45m. To make the crowds swoon, a new cart is brought in: a hollow sphere of mass 250kg and radius 2m, capable of carrying four passenger of total mass, 400kg.

The hollow sphere is capable of carrying passengers around the roller coaster separately. Friction although present is mostly negligible.

1.1 What is the initial energy of the system if the cart is released from rest?

1.2 What is the speed of cart when it reaches the bottom of the hill?

1.3 What is its speed at the top of the roller coaster loop?

1.4 What is the minimum speed to get around the loop? (Does it change or is it still the same?)

1.5 If the hollow sphere is replaced by a hollow disk with same mass and radius as the sphere, does the

speed in 1.4 change?

Answer 1

1.1

The mass of the hollow sphere and four passengers is

m = 250 kg + 400 kg

    = 650 kg

The corresponding figure is shown below:

It is given that the initial energy of the system if the cart is released from rest. Thus, the kinetic energy is zero. The total energy is entirely potential. It is given by

E = mgh

Here, acceleration due to gravity is g and height of the hill is h.

Substitute 650 kg for m, 125 m for h and 9.8 m/s² for g in the above equation,

E = mgh

   = (650 kg)(125 m)( 9.8 m/s²)

   = 7.9625 x 10⁵ J

Rounding off to three significant figures, the initial energy of the system is 7.96 x 10⁵ J.

1.2

At the bottom of the hill, the total energy is entirely kinetic as the potential energy is equal to zero.

Thus, the total energy at the bottom is

E₀ = ½ mv²

Here, speed of the system is v.

From the law of conservation of energy,

Substitute 125 m for h and 9.8 m/s² for g in the above equation,

Therefore, the speed of the system at the bottom of the hill is 35.0 m/s.

1.3

Let v₁ be the speed of the system at the top of the loop and R be the radius of the loop.

The total energy at the top of the loop is

As there is no loss of energy, the total energy is conserved.

From the law of conservation of energy,

Substitute 35 m/s for v, 9.8 m/s² for g and 2 m for R in the above equation,

Rounding off to three significant figures, the speed at the top of the loop is 23.9 m/s.

1.4

The minimum speed to get around the loop is

Substitute 9.8 m/s² for g and 2 m for R in the above equation,

Rounding off to three significant figures, the minimum speed to get around the loop is 4.43 m/s.

1.5

The minimum speed to get around the loop depends on the acceleration due to gravity and radius of the loop. It does not depend on the system in which it moves around the loop. Thus, it does have any effect whether the cart is the hollow sphere or hollow disk.

Therefore, the minimum speed remains the same if the hollow sphere is replaced by a hollow disk.