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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.

User Masroore
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2 Answers

3 votes

Answer:

Decreases by factor of 22

Step-by-step explanation:

khan acadamy

User Kris Gruttemeyer
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The expression for the angular acceleration is:

\alpha = (\Delta \omega)/(\Delta t)
where
\Delta \omega = \omega-\omega _0 is the variation of the angular velocity, with
\omega _0 being the starting velocity (which in our problem is zero), and
\Delta t being the time interval. So we can write the angular velocity after an angle
\delta \theta as

\omega (\delta \theta) = \alpha \Delta t
We also know the relationship between tangential velocity, v, and the angular velocity v:

v=\omega r
with r being the radius of the wheel. Substituting
\omega into the previous equation, we can write an expression for v:

v(\delta \theta )= \alpha r \Delta t
User Geekingreen
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