A uniform thin ring of mass $0.4$ kg rolls without slipping on a horizontal surface with a linear velocity of $10$ cm/s. The kinetic energy of the ring is?

4 \times 10^{- 3}

joules

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4 \times 10^{- 2}

joules

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2 \times 10^{- 3}

joules

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2 \times 10^{- 2}

joules

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Solution

The correct option is A $4 \times 10^{- 3}$ joules
${K.E}_{T o t a l} = {K.E}_{t r a n s l a t o r y} + {K.E}_{r o t a t i o n} {K.E}_{t r a n s l a t o r y} = \frac{1}{2} m v^{2} {K.E}_{r o t a t i o n} = \frac{1}{2} I ω^{2}$
Where moment of inertia of ring $= m r^{2}$
Reducing the above values we get
${K . E}_{T o t a l} = \frac{1}{2} m v^{2} + \frac{1}{2} m v^{2} = m v^{2}$
${K . E}_{T o t a l} = 0.4 \times (0.1)^{2} = 4 \times 10^{- 3} j o u l e s .$

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A uniform thin ring of mass 0.4 kg rolls without slipping on a horizontal surface with a linear velocity of 10 cm/s. The kinetic energy of the ring is?

The correct option is A 4×10−3 joules K.E Total= K.E translatory+ K.E rotation K.E translatory=12mv2 K.E rotation=12Iω2Where moment of inertia of ring=mr2Reducing the above values we getK.ETotal=12mv2+12mv2=mv2K.ETotal=0.4×(0.1)2=4×10−3joules.

A uniform thin ring of mass $0.4$ kg rolls without slipping on a horizontal surface with a linear velocity of $10$ cm/s. The kinetic energy of the ring is?