Question

A Ferris Wheel 22.0m in diameter rotates once every 12.5s. What is the ratio of a persons apperenet weight to her real weight (a.) at the top, and (b) at the bottom?

Answer 1

To find the ratio of a person's apparent weight to her real weight at the top and bottom of a Ferris wheel, use the formula F = m * (v^2 / r) where F is the centripetal force, m is the person's mass, v is the velocity of the Ferris wheel, and r is the radius of the Ferris wheel. At the top, the apparent weight increases, while at the bottom, the apparent weight decreases.

Step-by-step explanation:

To find the ratio of a person's apparent weight to her real weight at the top and bottom of the Ferris wheel, we need to consider the centripetal force acting on the person.

(a) At the top of the Ferris wheel, the person is experiencing an apparent weight that is equal to her real weight plus the centripetal force directed towards the center of the wheel. This adds to the person's apparent weight. The centripetal force can be calculated using the formula F = m * (v^2 / r), where m is the person's mass, v is the velocity of the Ferris wheel, and r is the radius of the Ferris wheel. At the top, the centripetal force is directed downwards which increases the apparent weight. The ratio of a person's apparent weight to her real weight at the top is the sum of the centripetal force and the person's real weight divided by her real weight.

(b) At the bottom of the Ferris wheel, the person is experiencing an apparent weight that is equal to her real weight minus the centripetal force directed towards the center of the wheel. This subtracts from the person's apparent weight. The centripetal force at the bottom is directed upwards which reduces the apparent weight. Using the same formula as above, the ratio of a person's apparent weight to her real weight at the bottom is the difference between the person's real weight and the centripetal force divided by her real weight.

Answer 2

Apparent weight: N = mg - F_up

So basically, you have two forces acting on the body - the force of gravity pulling it down, and the seat pushing it up.

The net force, F = m*a, and we're going to have to use centripetal acceleration in this problem, since it's a Ferris wheel.

So, centripetal acceleration is:

a = w^2 * r (w is omega in this case),

where w = 2pi/t

Ntop = m*(g - 2*Pi^2*d/T^2)
Nbottom = m*(g + 2*Pi^2*d/T^2)

We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground):

Ntop/(m*g) = m*(g - 2*Pi^2*d/T^2)/(m*g)
Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g)

Simplify:
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2)
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2)

Data:
d:=22 m; T:=12.5 sec; g:=9.8 N/kg;

Results:
Ntop/(m*g) = 71.64%.
Nbottom/(m*g) = 128.4%