Question

A flywheel flows from 250rpm to 150rpm in 4.2 seconds. How many revolutions occur during this time ​

Answer 1

The average speed during the slowdown is 200 RPM.

4.2 seconds is (4.2/60) = 0.07 minute.

(200 rev/minute)x(0.07 minute)= 14 revs

Answer 2

14 revolutions

Step-by-step explanation:

Given that,

Initial angular velocity,
`\omega_i=250\ rpm`

Final angular velocity,
`\omega_f=150\ rpm`

Time, t = 4.2 seconds

We need to find the number of revolutions occur durung this time.

250 rpm = 26.17 rad/s

150 rpm = 15.70 rad/s

Let
`\alpha` is angular acceleration. Using first equation to find it.

`\alpha =(\omega_f-\omega_i)/(t)\\\\\alpha =(15.70 -26.17 )/(4.2)\\\\\alpha =-2.49\ rad/s^2`

Now let us suppose that the number of revolutions are
`\theta`.

`\theta=(\omega_f^2+\omega_i^2)/(2\alpha)\\\\=(15.70^2-26.17^2)/(2* -2.49)\\\\=88\ rad`

or

`\theta=(88)/(2\pi)\\\\=14\ rev`

Hence, there are 14 revolutions.