Question

A flywheel of moment of inertia 9kgm is connected
to a motor. The motor accelerates from rest to
600 revolutions per minute. Neglecting friction
work done on the flywheel is ?

Answer 1

Answer:

Given:

Moment of Inertia of flywheel = 9 kgm²

Initial angular velocity = 0 rad/s

Final angular Velocity = 600 rpm

To find :

Work done on flywheel

Conversion:

Firstly convert all the quantities to SI units :

600 rpm = 600 × 2π/60 = 20π rad/s

Concept:

Work Done on the flywheel is equal to the change in angular Kinetic Energy as per

"Work-Energy Theorem"

Formulas used:

Angular Kinetic Energy = ½ × I × ω²,

where "I" is Moment of Inertia and ω is the angular velocity

Calculation:

Initial Angular Kinetic energy = 0 (as the flywheel is in rest)

Final Angular Kinetic energy

= ½ × I × ω²

= ½ × 9 × (20π)²

= ½ × 9 × 400 × π²

= 1800π² J

So work done = KE2 - KE1

=> W = 1800π² - 0

=> W = 1800π² Joules.

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A flywheel of moment of inertia 9kgm is connected to a motor. The motor accelerates from rest to 600 revolutions per minute. Neglecting friction work done on the flywheel is ?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Moment of Inertia (I) = 9 kg m².
• Frequency of Revolution(f) = 600 rev/min.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Here,

⇒ f = 600 rev/min.

Converting it to revolution/second.

Therefore,

$\large{\tt \leadsto f = \dfrac{600}{60} \: rev/sec}$

$\large{\tt \leadsto f = \cancel{\dfrac{600}{60}} \: rev/sec}$

$\large{\leadsto {\underline{\underline{\tt f = 10 \: rev/sec}}} \: \tt -----(1)}$

Now, Converting it into rad/sec(S.I units)

$\large{\boxed{\tt \omega = 2 \pi f}}$

Substituting the value of Frequency from Equation (1).

$\large{\tt \leadsto \omega = 2 \pi \times 10}$

$\large{\tt \leadsto \omega = 20 \times \pi}$

$\large{\leadsto{\underline{\underline{\tt \omega = 20 \pi}}} \: \tt ------(2)}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Rotational Kinetic Energy formula:-

$\large{\boxed{\tt K.E_r = \dfrac{1}{2} I \omega^2}}$

Substituting the values,

$\large{\tt \leadsto K.E_r = \dfrac{1}{2} \times 9 \times (20 \pi)^2}$

$\large{\tt \leadsto K.E_r = \dfrac{1}{2} \times 9 \times 400 \pi^2}$

$\large{\tt \leadsto K.E_r = \dfrac{1}{\cancel{2}} \times 9 \times \cancel{400} \pi^2}$

$\large{\tt \leadsto K.E_r = 9 \times 200 \pi^2}$

$\large{\boxed{\tt K.E_r = 1800 \pi^2 \: J}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

By Work - Energy theorem,

$\large{\boxed{\tt W = K.E_f - K.E_i}}$

Substituting the values,

$\large{\tt \leadsto W = 1800 \pi^2 - 0}$

As the Question specifies the initial angular velocity is zero.

Therefore,

$\huge{\boxed{\boxed{\tt W = 1800 \pi^2 \: J}}}$

So, the Work done by the Flywheel is 1800π² Joules.

$\rule{300}{1.5}$