Question

The diagram below shows a roller-coaster ride which contains a circular loop of radius r. A car (mass m) begins at rest from point A and moves down the frictionless track from A to B where it then enters the vertical loop (also frictionless), traveling once around the circle from B to C to D to E and back to B, after which it travels along the flat portion of The track from B to F (which is not frictionless).

Answer 1

The speed of the roller coaster at the top of the loop is 13.7 m/s.

Step-by-step explanation:

In order to calculate the speed of the roller coaster at the top of the loop, we need to consider the downward acceleration of the car and the radius of curvature of the loop. The downward acceleration of the car is given as 1.50 g, where g is the acceleration due to gravity. We can convert this to meters per second squared by multiplying it by 9.8 m/s². So, the downward acceleration of the car is 1.50 * 9.8 = 14.7 m/s².

The downward acceleration of the car at the top of the loop is due to the net force acting on the car, which is the difference between the centripetal force and the force due to gravity. The centripetal force is given by Fc = m * ac, where m is the mass of the car and ac is the centripetal acceleration. The centripetal acceleration is given by ac = v² / r, where v is the velocity of the car and r is the radius of curvature of the loop.

At the top of the loop, the downward centripetal acceleration should be greater than the acceleration due to gravity in order to keep the passengers pressed firmly into their seats. So, we can set up the following equation: m * ac - m * g = m * (v² / r) - m * 9.8 = 0. To solve for the velocity at the top of the loop, we rearrange the equation as follows: v² = r * 9.8, and take the square root of both sides to get the velocity: v = sqrt(r * 9.8). Plugging in the radius of curvature of 15.0 m, the velocity at the top of the loop is v = sqrt(15.0 * 9.8) = 13.7 m/s.