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?

Question

Mathematics

Freeman

6 January, 04:54

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?

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Answers (2)

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Javion Ramsey · Answer 1

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)

dа ta:

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

Results:

Ntop / (m*g) = 71.64%.

Nbottom / (m*g) = 128.4%

Tessa Ware · Answer 2

Acceleration of a passenger is centripetal acceleration, since the Ferris wheel is assumed at uniform speed:

a = omega^2*r

omega and r in terms of given dа ta:

omega = 2*Pi/T

r = d/2

Thus:

a = 2*Pi^2*d/T^2

What forces cause this acceleration for the passenger, at either top or bottom?

At top (acceleration is downward):

Weight (m*g) : downward

Normal force (Ntop) : upward

Thus Newton's 2nd law reads:

m*g - Ntop = m*a

At top (acceleration is upward):

Weight (m*g) : downward

Normal force (Nbottom) : upward

Thus Newton's 2nd law reads:

Nbottom - m*g = m*a

Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame:

Ntop = m * (g - a)

Nbottom = m * (g + a)

Substitute a:

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)

dа ta:

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

Results:

Ntop / (m*g) = 71.64% ... she feels "light"

Nbottom / (m*g) = 128.4% ... she feels "heavy"