Question

A flywheel slows from 600. to 416 rev/min while rotating through 34.0 revolutions. What is the angular acceleration of the flywheel

Answer 1

Answer:
`\alpha=4.798\ rad/s^2\quad [\text{deceleration}]`

Step-by-step explanation:

Given

Initial revolution of flywheel
`N_1=600\ rpm`

(Initial angular velocity
`\omega_i=(2\pi N_1)/(60)`)

Final revolution of flywheel
`N_2=600\ rpm`

(Final angular velocity
`\omega_f=(2\pi N_1)/(60)`)

Revolution turned
`34`

So, angle turned is
`\theta =2\pi * 34\\`

Using equation of angular motion i.e.
`\omega_f^2-\omega_i^2=2\cdot \alpha \cdot \theta`

`\Rightarrow \left((2\pi * 416)/(60)\right)^2-\left((2\pi * 416)/(60)\right)^2=2\alpha * (68\pi )\\\\\Rightarrow \alpha=(1898.344-3948.86)/(427.312)\\\\\Rightarrow \alpha =-4.798\ rad/s^2\\\Rightarrow \alpha =4.798\ rad/s^2\quad [\text{deceleration}]`