A thin metal disc of radius $25 c m$ and mass $2 k g$ starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is $8 J$ at the foot of the inclined plane, then the linear velocity of centre of mass of disc is :

2 m / s

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4 m / s

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6 m / s

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8 m / s

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Solution

The correct option is D $4 m / s$
$R = 0.25 m .$ (given)
$m = 2 k g .$ (given)
$K E)_{R} = 8 J .$
making the equation which makes the physical sense,as given below,
$8 = \frac{1}{2} \times (\frac{m}{2} R^{2}) {(\frac{V}{R})}^{2}$
$8 = \frac{m}{4} V^{2}$
$V = 4 m / s .$

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A thin metal disc of radius 25 cm and mass 2 kg starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is 8 J at the foot of the inclined plane, then the linear velocity of centre of mass of disc is :

The correct option is D 4 m/sR=0.25m. (given)m=2kg. (given)KE)R=8J.making the equation which makes the physical sense,as given below,8=12×(m2R2)(VR)28=m4V2V=4m/s.

A thin metal disc of radius $25 c m$ and mass $2 k g$ starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is $8 J$ at the foot of the inclined plane, then the linear velocity of centre of mass of disc is :

The correct option is D $4 m / s$
$R = 0.25 m .$ (given)
$m = 2 k g .$ (given)
$K E)_{R} = 8 J .$
making the equation which makes the physical sense,as given below,
$8 = \frac{1}{2} \times (\frac{m}{2} R^{2}) {(\frac{V}{R})}^{2}$
$8 = \frac{m}{4} V^{2}$
$V = 4 m / s .$