A circular solid disc of uniform thickness 20 mm, radius 200 mm and mass 20 kg, is used as aflywheel. If it rotates at 600 rpm, the kinetic energy of the flywheel, in Joules isa)395b)790c)1580d)3160Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Calculation of Kinetic Energy of the Flywheel

Given:
Thickness of the disc = 20 mm
Radius of the disc = 200 mm
Mass of the disc = 20 kg
Angular velocity = 600 rpm

To find:
Kinetic energy of the flywheel

Solution:

1. Calculation of Moment of Inertia (I)

The moment of inertia of a solid disc is given by the formula:

I = (1/2)mr²

where m = mass of the disc and r = radius of the disc

Substituting the given values, we get:

I = (1/2) x 20 x (0.2)²
I = 0.4 kgm²

2. Conversion of Angular Velocity to Radial Velocity

The angular velocity of the flywheel is given in rpm, which stands for revolutions per minute. To calculate the radial velocity, we need to convert it to radians per second.

1 revolution = 2π radians
1 minute = 60 seconds

Therefore, the angular velocity (ω) in radians per second is:

ω = 2π x (600/60)
ω = 62.83 rad/s

3. Calculation of Kinetic Energy (K)

The kinetic energy of the flywheel is given by the formula:

K = (1/2)Iω²

Substituting the values of I and ω, we get:

K = (1/2) x 0.4 x (62.83)²
K = 789.91 J

Therefore, the kinetic energy of the flywheel is 790 Joules (approx). The correct answer is option B.

Answer

Kinetic energy=1÷2×I×w×^2
I=1÷2×mass×radius^2
=1÷2×20×0.2^2m^2
=0.4
w=20pipersec
rotational kinetic energy=1÷2.I.w^2
=0.4kgm^2 20.pi^2
=788.8j

Answer

790j

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