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## Homework Statement

The figure below shows a Ferris wheel that rotates three times each minute. It carries each car around a circle of diameter 19.0 m.

a) What is the centripetal acceleration of a rider?

Answer in m/s^2

(b) What force does the seat exert on a 40.0 kg rider at the lowest point of the ride?

Answer in N

(c) What force does the seat exert on the rider at the highest point of the ride?

Answer in N

(d) What force (magnitude and direction) does the seat exert on a rider when the rider is halfway between top and bottom, going up?

Answer in N

Answer in ° (measured inward from the vertical)

## Homework Equations

## The Attempt at a Solution

a) Angular velocity w = 3 rev/min = 3 * 2pi/60 rad/s

= pi/10 rad/s

Or w = 0.314 rad/s

Radius r = 19.0 m/2 = 9.5 m

Centripetal acceleration a = w^2 * r

= 0.314^2 * 9.5

= 0.937 m/s^2

b) Mass M = 40.0 kg

Let force = F

F is upward and weight Mg is downward. Centripetal force is upward.

F - Mg = Ma

Or F = M(a+g) = 40 * (9.8 + 0.937) = 429.5 N

c) Mass M = 40.0 kg

Let force = F

F is upward and Mg is downward. Centripetal force is downward.

Mg - F = Ma

Or F = M(g - a) = 40 * (9.8 - 0.937) = 354 N

d) Let force = F at angle theta below vertical.

Vertical component of F = F cos(theta) upward.

Horizontal component of F = F sin(theta)

There is no acceleration in vertical direction.

Therefore F cos(theta) = Mg--------------------------(1)

Centripetal force is horizontal.

Therefore F sin(theta) = Ma---------------------------(2)

Dividing (2) by (1),

tan(theta) = a/g

Or theta = atan(a/g) = atan(0.937/9.8) = 5.46 deg

From (1),

F = Mg/cos(theta) = 40 * 9.8/cos(5.46 deg) = 394 N

Therefore, force by the seat is 394 N at 5.46 deg below vertical.

(Note: I do not know what exactly is meant by inward from vertical. I have calculated the angle from vertically upward. If you want the angle from vertically downward, then it is 180 - 5.46 = 174.54 deg

I only have one chance to enter this, please check to see if the way i did it is correct.