Question

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.46 rad/s2. Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of -31.4 rad/s. While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

Answer 1

The time for the change in the angular velocity to occur is 14.08 secs

Step-by-step explanation:

From the question,

the angular acceleration is - 4.46 rad/s²

Angular acceleration is given by the formula below

`\alpha =(\omega -\omega _(o) )/(t - t_(o) )`

Where
`\alpha` is the angular acceleration

`\omega` is the final angular velocity

`\omega _(o)` is the initial angular velocity

`t` is the final time

`t_(o)` is the initial time

From the question

`\alpha` = - 4.46 rad/s²

`\omega _(o)` = 0 rad/s (starting from rest)

`\omega` = -31.4 rad/s

`t_(o)` = 0 s

Now, we will determine t

From
`\alpha =(\omega -\omega _(o) )/(t - t_(o) )`, then

`-4.46 = (-31.4 - 0)/(t - 0)`

`-4.46 = (-31.4)/(t)`

`t = (-31.4)/(-4.46)`

t = 7.04 secs

This is the time spent in one direction,

Since the angular displacement of the wheel is zero ( it returned to its initial position), then the time required for the change in the angular velocity will be twice this time, that is 2t

Hence,

The time is 2×7.04 secs = 14.08 secs

This is the time for the change in the angular velocity to occur.