Question

A circular disc of mass 2 kg and radius 10 cm rolls without slipping with a speed 2 metre per second

Answer 1

ANSWER:-
Total kinetic energy of a rotating disk = rotational energy + Kinetic energy of the translation motion.

Kinetic energy = 1/2mv²

Rotational energy = Iω²/2

v = rω

ω = v/r

m = mass of the disk

v = linear velocity

I = moment of the disk

ω = angular velocity of the disk

I = 1/2mr²

Rotational energy = 1/2 × 1/2 × mr² × v²/r²

= 1/4mv²

Rotational energy is thus = 1/4mv²

Now doing the substitution :

K. E = 0.5 × 2 × 4 = 4 Joules

Rotational energy = 0.25 × 2 × 4 = 2 Joules

Total kinetic energy = 2 + 4 = 6

= 6J

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