Wheel of moment of Inertia 10kg meter square rotates 10 revolution per minute , find the work done in increasing its speed to 5 times of its initial value .
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Answered by
17
Hii dear,
# Answer- 394.78 J
# Given-
I = 30 kg.m^2
f = 10 rpm = 1/6Hz
ω2 = 5ω1
# Formula-
ω1 = 2π/T = 2π/6 = π/3 rad
KE1 = ½Iω1^2
KE1 = ½ × 30 × (π/3)^2
KE1 = 16.45 J
ω2 = 5ω1 = 5π/3 rad
KE2 = ½Iω2^2
KE1 = ½ × 30 × (5π/3)^2
KE2 = 411.23 J
KE1 = 16.45 J
Work done
= Change in KE
= KE2-KE1
= 411.23-16.45
= 394.78 J
Work done 394.78 Joule
# Answer- 394.78 J
# Given-
I = 30 kg.m^2
f = 10 rpm = 1/6Hz
ω2 = 5ω1
# Formula-
ω1 = 2π/T = 2π/6 = π/3 rad
KE1 = ½Iω1^2
KE1 = ½ × 30 × (π/3)^2
KE1 = 16.45 J
ω2 = 5ω1 = 5π/3 rad
KE2 = ½Iω2^2
KE1 = ½ × 30 × (5π/3)^2
KE2 = 411.23 J
KE1 = 16.45 J
Work done
= Change in KE
= KE2-KE1
= 411.23-16.45
= 394.78 J
Work done 394.78 Joule
Answered by
31
Answer: 131.46
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