Question

A heavy flywheel rotating on its central axis is slowing down because of friction in its bearings. At the end of the first minute of slowing, its angular speed is 0.68 of its initial angular speed of 260 rev/min. Assuming a constant angular acceleration, find its angular speed (rev/min) at the end of the second minute.

Answer 1

Answer:

426.787 rev/min

Explanation:

We know the normal equations of motion:

v = u + at

s = ut + $at^2/2$

$v^2-u^2=2as$

Now, likewise, in angular rotation,

u = ω$_o$ , where u: initial velocity, ω$_o$: initial angular velocity

v = ω , v: final velocity, ω: final angular velocity

a = α , a: acceleration, α: angular acceleration

t = t , t: time

s = θ , s: displacement, θ: angular displacement

Now, in this question:

First: ω$_o$ = 260 rev/min = 2π/60 (260) rad/s = 26π/3 rad/s

ω = 0.68(26π/3) rad/s , t = 60s

Let us find angular acceleration:

ω = ω$_o$ + αt

0.68(26π/3) = 26π/3 + 60α

α = 2.7733π/60

They want us to find angular speed at end of second minute, or t = 120s

Use:

ω = ω$_o$ + αt

ω = 26π/3 + 2.7733π/60 x 120

ω = 44.67 rad/s [Take π = 22/7]

To convert this into rev/minute:

ω = 44.67 x 60/2π = 426.787 rev/min

Hope it helps!