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13 days ago

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A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

heel starts turning, you are at the very top.Part AWhat is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘.Express your answer in degrees.Part BWhat is your speed v?

1 answer:

Maru [3.3K]13 days ago

6 0

A) 140 degrees

To initiate, we need to determine the angular velocity of the Ferris wheel. Given that its period is

T = 32 s

So, the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming that the wheel moves with a consistent angular velocity, we can now compute the angular displacement relative to the starting point:

$\theta= \omega t$

and substituting t = 75 seconds leads us to

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, this is

$15 rad: x = 2\pi rad: 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

Hence, the new position is 140 degrees from the initial reference position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the boundary of the wheel is calculated as

$v=\omega r$

where

$\omega=0.20 rad/s$

r = d/2 = (27 m)/2=13.5 m represents the radius of the wheel

Substituting into the equation, we will find

$v=(0.20 rad/s)(13.5 m)=2.7 m/s$

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