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प्रश्न
A uniform disc of mass 10 kg and radius 60 cm rotates about an axis perpendicular to its plane and passing through its centre at 1200 rpm. Calculate its rotation kinetic energy. [Take π2 = 10]
संख्यात्मक
उत्तर
Data: M = 10 kg, R = 0.6 cm, f = 1200 rpm = `1200/(60 Hz)` = 20 Hz
`I_(disc) = 1/2 MR^2`
Rotational kinetic energy = `1/2 I_(disc) ω^2`
= `1/2 (1/2 MR^2) (2πf)^2`
= `1/2 Mπ^2 (Rf)^2`
= `1/2 (10) (10) (0.6 xx 20)^2`
= `1/2 (100) (144)`
= 7200 J
= 7.2 kJ
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