Question

A circular disc of mass 0.5 kg and radius 0.1 m is making 60 revolution/min about an axis passing through its centre and perpendicular to its plane. Calculate its kinetic energy

Answer 1

A circular disc of mass 0.5 kg and radius 0.1 m is making 60 revolution/min about an axis passing through its centre and perpendicular to its plane. Calculate its kinetic energy

Mass = 0.5 kg

Radius = 0.1 m

w = 60 revolution / min

=> 60× 2π/60 rad / sec

=> 2π rad / sec

Formula used :

I = MR² /2

=> 0.0025

_________________

K.E = 1/2 I W²

= 0.00785 J

Answer 2

Answer:

A circular disc of mass 0.5 kg and radius 0.1 m is making 60 revolutions/min about an axis passing through its center and perpendicular to its plane. Its kinetic energy is 0.147894 J.

Explanation:

Given,

Mass of the disc   (M) = 0.5 kg

Radius of the disc (R)  =  0.1 m

Angular velocity  $\omega=60 R P M=\frac{60 \times 2 \pi}{60}=2 \pi \mathrm{Rad} / \mathrm{sec}$

For a circular disc, the moment of inertia is given by,

I  $=\frac{3}{2} M R^{2}$  (Using theorem of  parallel axes)

I  $=\frac{3}{2}\times 0.5 \times 0.1^{2}$

  = 0.0075

Rotational kinetic energy can be written as,

$K.E_{rotational} =\frac{1}{2} I \omega^{2}$

Substituting the values of I and ω,

$K.E_{rotational} =\frac{1}{2} \times 0.0075\times \ (2 \pi )^{2}$

                     = 0.147894 J

Ans:

Kinetic energy = 0.147894 J