a thin ring of mass 1 kg and radius 1m is rolling at a speed of 1ms .its kinetic energy is
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Linear kinetic energy = 0.5mv²
Rotational kinetic energy = 0.5mv²k²/r²
Let k²/r² = β
Rotational kinetic energy = 0.5mv²β
Total Kinetic energy = Linear kinetic energy + Rotational kinetic energy
= 0.5mv²[1 + β]
= 0.5 × 1 kg × (1 m/s)²[1 + 1] ………[∵ k²/r² value of ring about its central axis is 1]
= 1 Joule
Total Kinetic energy of ring is 1 joule
Rotational kinetic energy = 0.5mv²k²/r²
Let k²/r² = β
Rotational kinetic energy = 0.5mv²β
Total Kinetic energy = Linear kinetic energy + Rotational kinetic energy
= 0.5mv²[1 + β]
= 0.5 × 1 kg × (1 m/s)²[1 + 1] ………[∵ k²/r² value of ring about its central axis is 1]
= 1 Joule
Total Kinetic energy of ring is 1 joule
sudhanshu1624:
really thank you
You’re welcome :)
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