An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :

Question

An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is : (A) 0.7 kg - m 2 (B) 3.22 kg - m 2 (C) 30.8 kg - m 2 (D) 0.07 kg - m 2

Tardigrade Admin · Accepted Answer

ω i =60 rpm =60 × (2 π/60)=2 π rad / s ω f =360 rpm =360 × (2 π/60)=12 π rad / s Δ K . E .=(1/2) I (ω f 2-ω i 2)=484 (1/2) I (144 π2-4 π2)=484 I ≈ 0.7 kg - m 2

Q. An energy of $484 J$ is spent in increasing the speed of a flywheel from $60 r p m$ to $360 r p m$ . The moment of inertia of the flywheel is :

$0.7 k g - m^{2}$

$3.22 k g - m^{2}$

$30.8 k g - m^{2}$

$0.07 k g - m^{2}$

$ω_{i} = 60 r p m = 60 \times \frac{2 π}{60} = 2 π r a d / s$
$ω_{f} = 360 r p m = 360 \times \frac{2 π}{60} = 12 π r a d / s$
$Δ K . E . = \frac{1}{2} I (ω_{f}^{2} - ω_{i}^{2}) = 484$
$\frac{1}{2} I (144 π^{2} - 4 π^{2}) = 484$
$I \approx 0.7 k g - m^{2}$

Q. An energy of 484J is spent in increasing the speed of a flywheel from 60rpm to 360rpm. The moment of inertia of the flywheel is :

0.7kg−m2

3.22kg−m2

30.8kg−m2

0.07kg−m2