The moment of inertia of a body about a given axis is 1.2kgm^2 . Initially the body is at rest. In order to produce a rotational kinetic energy of 1500J, an angular acceleration of 25rad/second ^2 must be applied about that axis for a duration of

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Monali Verma

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Rotation kinetic energy= ½Iω² Where I = moment of Inertia ω = angular velocity And, kinematic relation for constant angular acceleration is ω(final) = ω(initial) + αt Where, α = angular acceleration t = time
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Rotation kinetic energy= ½Iω² Where I = moment of Inertia ω = angular velocity And, kinematic relation for constant angular acceleration is ω(final) =ω(initial) +αt Where,α = angular acceleration t = time read less
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solution to above question
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Answer : K= rotational kinetic energy ω = angular velocity I = moment of inertia
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KE = Iw²/2 1500=1.2×w²/2 ω=50 ωo=0 ω=ωo+αt 25t=50 t= 2sec ans
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KE = Iw²/2 1500=1.2×w²/2 ω=50 ωo=0 ω=ωo+αt 25t=50 t= 2sec ans read less
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Msc physics student at a central university

t=1500/(1.2*625) This is straight from the formula and you get 2 seconds
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The correct option is B 2 sI=1.2 kgm2;t=?Krot=1500 Jα=25rad/s2ω=ω0+αtInitially the body was at rest, ωo=0Rotational kinetic energy, K=12Iω2⇒1500=12×1.2×w2w=√30001.2=50 rad/sω=ω0+αtωo=0t=wα=5025=2s
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The correct option isB2 sI=1.2 kgm2;t=?Krot=1500 Jα=25rad/s2ω=ω0+αtInitially the body was at rest,ωo=0Rotational kinetic energy,K=12Iω2⇒1500=12×1.2×w2w=√30001.2=50 rad/sω=ω0+αtωo=0t=wα=5025=2s read less
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Masters in Physics from NIT Jaipur

t=1500/(1.2*625) sec. Solvig this you get t=2sec
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15 years of experience in science and mathematics

Hello you can call me i willl explain it
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