The radius of curvature of a loop-the-loop for a roller coaster is 11.8 m. At the top of the loop (with the car inside the loop), the force that the seat exerts on a passenger of mass m is 0.80mg. Find the speed of the roller coaster at the top of the loop.
To find the speed of the roller coaster at the top of the loop, we can use the concept of centripetal force and gravitational force.
1. Start by identifying the forces acting on the passenger at the top of the loop:
- Gravitational force (mg) acting downward.
- Normal force (the force exerted by the seat) acting towards the center of the loop, providing the centripetal force.
2. At the top of the loop, the net force acting on the passenger is the difference between the centripetal force and gravitational force:
Net force = Centripetal force - Gravitational force
3. The centripetal force is given by the normal force because it provides the necessary force to keep the passenger moving in a circular path:
Centripetal force = Normal force
4. Use the given information that the force the seat exerts on the passenger is 0.80mg, we can set up the equation for the net force:
Net force = 0.80mg - mg
5. Simplify the equation:
Net force = 0.80mg - mg
Net force = 0.20mg
6. At the top of the loop, the net force is given by the equation:
Net force = (mass) x (centripetal acceleration)
Centripetal acceleration can be calculated using the formula:
Centripetal acceleration = (velocity^2) / (radius of curvature)
7. Substitute the values into the net force equation:
0.20mg = (m * (v^2)) / r
8. Rearrange the equation to isolate the velocity (v):
v^2 = (0.20mg * r) / m
v^2 = 0.20g * r
9. Take the square root of both sides to solve for the velocity:
v = sqrt(0.20g * r)
10. Use the given radius of curvature (11.8 m) and the acceleration due to gravity (9.8 m/s^2) to find the speed:
v = sqrt(0.20 * 9.8 * 11.8)
v = sqrt(23.104)
v = 4.807 m/s
Therefore, the speed of the roller coaster at the top of the loop is approximately 4.807 m/s.