(a) The centripetal acceleration (a) can be calculated using the formula a = ω²r, where ω is the angular velocity and r is the radius of the Ferris wheel.

First, we need to convert the rotation speed from rotations per minute to radians per second. There are 2π radians in one rotation and 60 seconds in one minute, so ω = 4 rotations/minute * 2π rad/rotation / 60 s/minute = 0.4189 rad/s.

The radius of the Ferris wheel is half its diameter, so r = 22.0 m / 2 = 11.0 m.

Substituting these values into the formula gives a = (0.4189 rad/s)² * 11.0 m = 1.93 m/s². The direction of the centripetal acceleration is always towards the center of the circle, so in this case it is downwards at the highest point of the ride, upwards at the lowest point, and horizontal at the points halfway between the top and bottom.

(b) The force exerted by the seat on the child at the lowest point of the ride can be calculated using the formula F = m(a + g), where m is the mass of the child, a is the centripetal acceleration, and g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the given values gives F = 42.0 kg * (1.93 m/s² + 9.8 m/s²) = 494.6 N. The direction of this force is upwards, because the seat is pushing the child up against the downward force of gravity.

(c) At the highest point of the ride, the force exerted by the seat on the child is F = m(g - a), because the seat is now pushing the child down against the upward force of the centripetal acceleration.

Substituting the given values gives F = 42.0 kg * (9.8 m/s² - 1.93 m/s²) = 331.4 N. The direction of this force is downwards.

(d) When the child is halfway between the top and bottom and moving upward, the force exerted by the seat on the child is the same as at the highest point, because the centripetal acceleration is still directed towards the center of the Ferris wheel and the force of gravity is still acting downwards. Therefore, F = 331.4 N.

The direction of this force is at an angle θ below the horizontal, where θ can be calculated using the formula θ = arctan(a/g). Substituting the given values gives θ = arctan(1.93 m/s² / 9.8 m/s²) = 11.2°. Because the child is moving upward, this angle is counter-clockwise from the horizontal.

Question

(a) The centripetal acceleration (a) can be calculated using the formula a = ω²r, where ω is the angular velocity and r is the radius of the Ferris wheel.

First, we need to convert the rotation speed from rotations per minute to radians per second. There are 2π radians in one rotation and 60 seconds in one minute, so ω = 4 rotations/minute * 2π rad/rotation / 60 s/minute = 0.4189 rad/s.

The radius of the Ferris wheel is half its diameter, so r = 22.0 m / 2 = 11.0 m.

Substituting these values into the formula gives a = (0.4189 rad/s)² * 11.0 m = 1.93 m/s². The direction of the centripetal acceleration is always towards the center of the circle, so in this case it is downwards at the highest point of the ride, upwards at the lowest point, and horizontal at the points halfway between the top and bottom.

(b) The force exerted by the seat on the child at the lowest point of the ride can be calculated using the formula F = m(a + g), where m is the mass of the child, a is the centripetal acceleration, and g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the given values gives F = 42.0 kg * (1.93 m/s² + 9.8 m/s²) = 494.6 N. The direction of this force is upwards, because the seat is pushing the child up against the downward force of gravity.

(c) At the highest point of the ride, the force exerted by the seat on the child is F = m(g - a), because the seat is now pushing the child down against the upward force of the centripetal acceleration.

Substituting the given values gives F = 42.0 kg * (9.8 m/s² - 1.93 m/s²) = 331.4 N. The direction of this force is downwards.

(d) When the child is halfway between the top and bottom and moving upward, the force exerted by the seat on the child is the same as at the highest point, because the centripetal acceleration is still directed towards the center of the Ferris wheel and the force of gravity is still acting downwards. Therefore, F = 331.4 N.

The direction of this force is at an angle θ below the horizontal, where θ can be calculated using the formula θ = arctan(a/g). Substituting the given values gives θ = arctan(1.93 m/s² / 9.8 m/s²) = 11.2°. Because the child is moving upward, this angle is counter-clockwise from the horizontal.

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