A thin metal disc of radius $0.25 m$ and mass $2 k g$ starts from rest and rolls down an inclined plane. If its rotational kinetic energy is $4 J$ at the foot of the inclined plane, then its linear velocity at the same point is:

1.2 m / s^{1}

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2.8 m / s^{1}

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20 m / s^{1}

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2 m / s^{1}

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Solution

The correct option is B $2.8 m / s^{1}$
$K E_{R} = \frac{1}{2} I ω^{2} = 4 J$

$\frac{1}{2} \times \frac{1}{2} m r^{2} {(\frac{v}{r})}^{2} = 4 J$

$\frac{1}{4} m v^{2} = 4 J$

$v^{2} = \frac{4 \times 2 \times 2}{2}$

$v = 2 \sqrt{2} = 2.82 m / s$

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A thin metal disc of radius 0.25 m and mass 2 kg starts from rest and rolls down an inclined plane. If its rotational kinetic energy is 4 J at the foot of the inclined plane, then its linear velocity at the same point is:

The correct option is B 2.8 m/s1KER=12Iω2=4J12×12mr2(vr)2=4J14mv2=4Jv2=4×2×22v=2√2=2.82m/s

A thin metal disc of radius $0.25 m$ and mass $2 k g$ starts from rest and rolls down an inclined plane. If its rotational kinetic energy is $4 J$ at the foot of the inclined plane, then its linear velocity at the same point is:

The correct option is B $2.8 m / s^{1}$
$K E_{R} = \frac{1}{2} I ω^{2} = 4 J$

$\frac{1}{2} \times \frac{1}{2} m r^{2} {(\frac{v}{r})}^{2} = 4 J$

$\frac{1}{4} m v^{2} = 4 J$

$v^{2} = \frac{4 \times 2 \times 2}{2}$

$v = 2 \sqrt{2} = 2.82 m / s$