Question

A ring of radius 20 cm and mass 5 kg rolls on a smooth horizontal surface at the rate of 2 m/s. Rotational kinetic energy of a ring is
10 J
20 J
30 J
40 J

Answer 1

Answer:

Explanation:

ROTATIONAL KINETIC ENERGY = 1/2 I ω²

  MOMENT OF INERTIA FOR RING = MR²        (RADIUS = 20 cm = 0.2 m)

                                   =  5 * 0.2 = 1 kgm²

           ANGULAR VELOCITY ω = V / R ( V IS THE LINEAR VELOCITY)

                                                 ω = 2/0.2 = 10 rad/s

ROTATIONAL KINETIC ENERGY =1/2 * 1 * 10²

                                                      = 50 J

Answer 2

${\underline{\sf{Given}}}$

Radius of ring = 20cm = 0.2m

Mass of ring = 5kg

It rolls on a smooth surface at the rate = 2m/s

${\underline{\sf{To\:Find}}}$

Rotational kinetic energy of a ring

${\underline{\sf{Solution}}}$

Let consider that ring is rotated about an Axis pperpendicular to plane passing through the centre of mass.

⇒ I = MR²

We know that Kinteic energy of a rolling body is :

$\sf{K.E.=\dfrac{1}{2}mv{}^{2}+\dfrac{1}{2}I\omega}$

$\sf{\implies \frac{1}{2} \times 5 \times {2}^{2} + \frac{1}{2} \times 5 \times {0.2}^{2} \times \frac{{2}^{2}}{{0.2}^{2}}}$

$\sf{\implies 10 + \frac{1}{2} 5 \times 0.04 \times \frac{400}{4} }$

$\sf{\implies 10 + 5 \times 0.02 \times 100}$

${\boxed{\sf{\implies 10 + 10 = 20 \; J}}}$

More About rolling motion

When a body performs translatory motion as well as rotatory motion then it is known as rolling motion.