MechanicalDynamicsmediumFE ExamCoursework

Revolutions During Flywheel Run-Down

A flywheel spinning at $1200\ \text{rpm}$ is brought uniformly to rest in $40\ \text{s}$ .

How many complete revolutions does it turn through while stopping?

Given

N0	1200 rpm	Initial speed
t	40 s	Time to stop

Hint 1

Rotational kinematics mirrors linear kinematics — θ, ω, α replace s, v, a.

Hint 2

Convert rpm to rad/s first, then find α from the uniform stop.

Hint 3

Shortcut: for constant deceleration, average speed is half the initial speed.

Worked solution — try the problem first

Step 1 — Convert to rad/s.

$\omega_0 = \frac{2\pi N}{60} = \frac{2\pi(1200)}{60} = 125.66\ \text{rad/s}$

Step 2 — Angular deceleration. Uniform, from $\omega_0$ to zero in 40 s:

$\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 125.66}{40} = -3.1416\ \text{rad/s}^2$

Step 3 — Angular displacement.

$\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 = 125.66(40) + \tfrac{1}{2}(-3.1416)(1600)$ $= 5026.5 - 2513.3 = 2513.3\ \text{rad}$

Step 4 — Convert to revolutions.

$n = \frac{2513.3}{2\pi} = 400\ \text{rev}$

$\boxed{400\ \text{revolutions}}$

Faster route. For constant deceleration the average speed is half the initial, so

$n = \frac{1200\ \text{rpm}}{2} \times \frac{40}{60}\ \text{min} = 600 \times 0.6667 = 400\ \text{rev} \ \checkmark$

Concepts:Rotational kinematicsAngular accelerationUnit conversion

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