Question

A flywheel in the form of disc is rotating about
an axis passing through its centre and
perpendicular to its plane looses 100J of energy,
when slowing down from 60 r.p.m. to 30 r.p.m.
Find its moment of inertia about the same axis
and change in its angular momentum.​

Answer 1

Hello Students,

◆ Answer -

I = 6.755 kgm²

∆L = -21.22 kgm²/s

◆ Explanation -

# Given -

∆KE = 100 J

f1 = 60 rpm = 1 Hz

f2 = 30 rpm = 0.5 Hz

# Solution -

Initial kinetic energy is -

KEi = 1/2 Iw1²

KEi = 1/2 × I × (2π×1)²

KEi = 2π²I

Final kinetic energy is -

KEf = 1/2 Iw2²

KEf = 1/2 × I × (2π×0.5)²

KEf = π²I/2

Change in kinetic energy of flywheel is given as 100 J.

∆KE = KEf - KEi

100 = 2π²I - π²I/2

I = 100 × 2 / (3 × π²)

I = 6.755 kgm²

Change in angular momentum is -

∆L = Iw2 - Iw1

∆L = 2πI (f2 - f1)

∆L = 2π × 6.755 (0.5 - 1)

∆L = -21.22 kgm²/s

Hope this helps you.

Answer 2

Answer:

◆ Answer -

I = 6.755 kgm²

∆L = -21.22 kgm²/s

◆ Explanation -

# Given -

∆KE = 100 J

f1 = 60 rpm = 1 Hz

f2 = 30 rpm = 0.5 Hz

# Solution -

Initial kinetic energy is -

KEi = 1/2 Iw1²

KEi = 1/2 × I × (2π×1)²

KEi = 2π²I

Final kinetic energy is -

KEf = 1/2 Iw2²

KEf = 1/2 × I × (2π×0.5)²

KEf = π²I/2

Change in kinetic energy of flywheel is given as 100 J.

∆KE = KEf - KEi

100 = 2π²I - π²I/2

I = 100 × 2 / (3 × π²)

I = 6.755 kgm²

Change in angular momentum is -

∆L = Iw2 - Iw1

∆L = 2πI (f2 - f1)

∆L = 2π × 6.755 (0.5 - 1)

∆L = -21.22 kgm²/s