Question

You have to design a loop-the-loop for a new amusement park so that when each car passes the top of the loop inverted (upside-down), each seat exerts a force against a passenger's bottom that has a magnitude equal to 1.5 times the gravitational force that Earth exerts on the passenger.

Write the equation for the speed of car on the top of the loop-the-loop for a loop of radius R. Assume the mass of car is m.

Answer 1

The equation for the speed of the car on the top of the loop-the-loop is v = sqrt(2.5 * g * R), where g is the acceleration due to gravity and R is the radius of the loop.

Step-by-step explanation:

To find the speed of the car at the top of the loop-the-loop, we can use the concept of centripetal force. At the top of the loop, the gravitational force and the normal force (exerted by the seat on the passenger) provide the centripetal force. The formula for centripetal force is:

F = m * a

where F is the net force, m is the mass, and a is the centripetal acceleration. In this case, the net force is the sum of the gravitational force (m * g, where g is the acceleration due to gravity) and the normal force (1.5 * m * g). Equating the net force to the centripetal force (m * a), we can solve for the acceleration:

m * a = m * g + 1.5 * m * g

Simplifying the equation, we have:

a = 2.5 * g

Since the centripetal acceleration is equal to the acceleration due to gravity at the top of the loop, we can use the formula for centripetal acceleration to find the speed:

a = v^2 / R

where v is the speed and R is the radius of the loop. Substituting the value of centripetal acceleration, we get:

2.5 * g = v^2 / R

Solving for v, we find:

v = sqrt(2.5 * g * R)

Therefore, the equation for the speed of the car on the top of the loop is v = sqrt(2.5 * g * R), where g is the acceleration due to gravity and R is the radius of the loop.

Answer 2

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