Chapter 31 : Kinetics of Motion of Rotation „„„„„ 631

We know that maximum speed of the wheel

N 1 = 360 + ( 0.02 × 360) = 367.2 r.p.m.

and minimum speed of the wheel,

N 2 = 360 – (0.02 × 360) = 352.8 r.p.m.

∴ Fluctuation of energy (E),

30 000 =

2

900 IN N(– )^12 N

π

×

2

360 (367.2 – 352 8). 56.85

900

II

π

=×× =

∴ I=^2

30 000

527.7 kg-m

56.85

= Ans.

(ii) Mass of the wheel

We know that mass of the wheel,

M= 22

527.7

651.5 kg

(0.9)

I

k

== Ans.

Example 31.4. A flywheel of mass 8 tonnes starts from rest, and gets up a speed of 180
r.p.m. in 3 minutes. Find the average torque exerted on it, if the radius of gyration of the flywheel is
60 cm.

Solution. Given: Mass of the flywheel (M) = 8 t = 8000 kg; Initial angular speed (ω 0 ) = 0

(because it starts from rest); Final angular speed (ω) = 180 r.p.m. =

180 2

60

×π

= 6π rad/s.; Time

(t) = 3 min = 180 s and radius of the gyration of the flywheel (k) = 60 cm = 0.6 m.

Let α= Constant angular acceleration of the flywheel

We know that the mass moment of inertia of the flywheel,

I= Mk^2 = 8000 × (0.6)^2 = 2880 kg-m^2

and final angular velocity of the flywheel (ω),

6 π=ω 0 + αt = 0 + α × 180

or α=

(^6) 0.105 rad/s 2
180
π

∴ Average torque exerted by the flywheel.
T=Iα = 2880 × 0.105 = 302.4 N-m Ans.
Example 31.5. A flywheel is made up of steel ring 40 mm thick and 200 mm wide plate
with mean diameter of 2 metres. If initially the flywheel is rotating at 300 r.p.m., find the time taken
by the wheel in coming to rest due to frictional couple of 100 N-m.
Take mass density of the steel as 7900 kg/m^3. Neglect the effect of the spokes.
Solution. Given: Thickness of flywheel = 40 mm = 0.04 m; Width of flywheel = 200 mm
= 0.2 m; Mean diameter of flywheel = 2 m or mean radius (r) = 1 m; Initial angular speed
(ω 0 ) = 300 r.p.m. = 5 r.p.s. = 10π rad/s; Frictional couple = 100 N-m and density of steel = 7900 kg/m^3.
Let α= Constant angular acceleration of flywheel and
t= Time taken by the flywheel in coming to rest.
We know that volume of flywheel,
=π × 2 × 0.2 × 0.04 = 0.05 m^3
∴ Mass of the flywheel,
M= 0.05 × 7900 = 395 kg