Question

The model of a roller coaster consists of a ramp and a circular loops. A small block of mass M starts from heigh H with speed V=0 and slides without friction down the ramp and around the inside of the loop of radius H/2. Position P is a point on the loop at an angle of θ with the horizontal. express all algebraic answers in terms of M,H,θ and fundamental constants. Calculate the angle θL in degrees at which the block leaves the loop

Answer 1

The speed of the roller coaster at the top of the loop can be calculated using the conservation of energy. It is given by v = sqrt(3gh), where v is the velocity, g is the acceleration due to gravity, and h is the height from the ground.

Step-by-step explanation:

The speed of the roller coaster at the top of the loop can be calculated using the conservation of energy. At the top of the loop, the roller coaster will have both kinetic and potential energy. The potential energy is given by the height from the ground, which is equal to the radius of the loop (H/2). The kinetic energy is given by the formula KE = 0.5mv^2, where m is the mass of the block and v is the velocity at the top of the loop.

Since there is no friction, the total mechanical energy at the start of the coaster is equal to the total mechanical energy at the top of the loop:

1. At the start: mgh
2. At the top: 0.5mv^2 + mgh/2

Setting these two equations equal to each other and solving for v will give us the velocity of the roller coaster at the top of the loop:

mgh = 0.5mv^2 + mgh/2

v = sqrt(3gh)