A ferris wheel of radius r speeds up with angular acceleration α starting from rest. part a find an expression for the velocity of a rider after the ferris wheel has rotated through angle δθ.

Respuesta :

The expression for the angular acceleration is:
[tex]\alpha = \frac{\Delta \omega}{\Delta t} [/tex]
where [tex]\Delta \omega = \omega-\omega _0[/tex] is the variation of the angular velocity, with [tex]\omega _0[/tex] being the starting velocity (which in our problem is zero), and [tex]\Delta t[/tex] being the time interval. So we can write the angular velocity after an angle [tex]\delta \theta[/tex] as
[tex]\omega (\delta \theta) = \alpha \Delta t[/tex]
We also know the relationship between tangential velocity, v, and the angular velocity v:
[tex]v=\omega r[/tex]
with r being the radius of the wheel. Substituting [tex]\omega[/tex] into the previous equation, we can write an expression for v:
[tex]v(\delta \theta )= \alpha r \Delta t [/tex]

Answer:

Decreases by factor of 22

Explanation:

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