Question

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.

Answer 1

Decreases by factor of 22

Step-by-step explanation:

khan acadamy

Answer 2

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`