An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5.1 kN, and the circle’s radius is 10 m. At the top of the circle, what are the (a) magnitude FB and (b) direction (up or down) of the force on the car from the boom if the car’s speed is v= 4.0 m/s? What are the magnitude(c) FB and (d) the direction (up or down) of the force if the cars speed is v= 13m/s?
To find the magnitude and direction of the force on the car from the boom, we can consider the forces acting on the car at the top of the circle.
(a) Finding the magnitude of the force FB with a car speed v = 4.0 m/s:
At the top of the circle, the car is moving in a horizontal direction, so the net force acting on the car must be directed towards the center of the circle to provide the necessary centripetal force. The net force is the vector sum of the gravitational force (mg) and the force from the boom (FB).
Using Newton's second law, we can write the equation:
mg + FB = mv²/r
Given:
m = mass = combined weight/g = 5.1 kN / 9.8 m/s² = 5.1 × 10³ kg
v = 4.0 m/s
r = radius = 10 m
Plugging in the values:
(5.1 × 10³ kg) × (9.8 m/s²) + FB = (5.1 × 10³ kg) × (4.0 m/s)² / 10 m
Simplifying the equation and solving for FB:
FB = (5.1 × 10³ kg) × [ (4.0 m/s)² / 10 m - 9.8 m/s² ]
FB ≈ (5.1 × 10³ kg) × (16 m²/s² / 10 m - 9.8 m/s² )
FB ≈ 8160 N - 4998 N = 3162 N
Therefore, the magnitude of the force FB on the car from the boom when the car's speed is 4.0 m/s is approximately 3162 N.
(b) Determining the direction:
Since the force FB is positive (3162 N) and directed upward, we can conclude that the direction of the force on the car from the boom is upward.
(c) Finding the magnitude of the force FB with a car speed v = 13 m/s:
Using the same method as above, we can calculate the magnitude of the force FB for a car speed of 13 m/s.
Given:
m = 5.1 × 10³ kg
v = 13 m/s
r = 10 m
Plugging in the values and solving for FB:
FB = (5.1 × 10³ kg) × [ (13 m/s)² / 10 m - 9.8 m/s² ]
FB ≈ (5.1 × 10³ kg) × (16.9 m²/s² / 10 m - 9.8 m/s² )
FB ≈ 8624.5 N - 4998 N = 3626.5 N
Therefore, the magnitude of the force FB on the car from the boom when the car's speed is 13 m/s is approximately 3626.5 N.
(d) Determining the direction:
Similar to (b), since the force FB is positive (3626.5 N) and directed upward, we can conclude that the direction of the force on the car from the boom is upward.