Question

The angular velocity of flywheel decreases uniformly from 12000rev/min to 3000 rev/min in 8 sec. Find the angular acceleration and the number of revolutions made by the wheel in the 8 sec interval. How many more seconds are required for the wheel to come to rest?

Answer 1

Initial rotational speed of the flywheel = N₁ = 12000 rev/min

Rotational speed of the flywheel after 8 sec = N₂ = 3000 rev/min

Initial angular speed of the flywheel = $\omega$₁

$\omega$₁ = 2$\pi$N₁/60

$\omega$₁ = 2$\pi$(12000)/60

$\omega$₁ = 1256.637 rad/s

Angular speed of the flywheel after 8 sec= $\omega$₂

$\omega$₂ = 2$\pi$N₂/60

$\omega$₂ = 2$\pi$(3000)/60

$\omega$₂ = 314.16 rad/s

Angular acceleration of the flywheel = $\alpha$

Time period = t₁ = 8 sec

$\omega$₂ = $\omega$₁ + $\alpha$t₁

314.16 = 1256.637 + $\alpha$(8)

$\alpha$ = -117.81 rad/s²

Negative sign as it is deceleration.

Angle turned by the flywheel in the 8 sec = $\theta$

$\omega$₂² = $\omega$₁² + 2$\alpha$$\theta$

(314.16)² = (1256.637)² + 2(-117.81)$\theta$

$\theta$ = 6283.168 rad

Number of revolutions made by the flywheel in the 8 sec = n

n = $\theta$/(2$\pi$)

n = 6283.168/(2$\pi$)

n = 1000 rev

Additional time required for the wheel to come to rest = t₂

Final angular speed of the flywheel = $\omega$₃ = 0 rad/s (comes to rest)

$\omega$₃ = $\omega$₂ + $\alpha$t₂

0 = 314.16 + (-117.81)t₂

t₂ = 2.67 sec

Angular acceleration of the flywheel = -117.81 rad/s²

Number of revolutions made by the wheel in 8 sec = 1000 rev

Additional time required for the wheel to come to rest = 2.67 sec