Question

A flywheel rotating at 420 rpm slows down at a constant rate of 2 rad/s^2. What time is required to stop the flywheel.

Answer 1

Final Answer : $t = 7 \pi s$  

Steps:

1) We have,  

$\omega_i =420rpm= 14 \pi rad/s \\</p><p>\omega_f =0 \\  \alpha = -2 rad/s^2$  

2) Then,  we know

 $\omega_f = \omega_i +\alpha t \\  => 0= 140 \pi - \alpha t \\  => t = 7 \pi$  

At t =7 π s,  flywheel will stop.

Answer 2

Hii Friend,

◆ Answer-
t = 22 s

◆ Explaination-
# Given-
α = -2 rad/s^2
f = 420 rpm = 7 Hz
ω2 = 0 rad/s

# Solution-
Initially angular velocity of the flywheel is-
ω1 = 2πf
ω1 = 2×3.14×7
ω1 = 44 rad/s

Using laws for rotational motion-
ω2 = ω1 + αt
0 = 44 + (-2)t
t = 44/2
t = 22 s

Therefore, the flywheel will take 22 s to stop.

Hope this helps you...