Question

calculate the K.E of rotation of a circular disc of mass 1kg and radius 0.2m rotating about an axis passing through the centre and perpendicular to it's plane .The disc makes 30/π rotations per minute​

Answer 1

Explanation:

fine here w is angular velocity . where n is no of rotations per second but in question it is given that in minute so convert into second and proceed okk.

Answer 2

Given : Mass of circular disc = 1 kg

            Radius of circular disc = 0.2 m

            Disc makes 30/π rotations per minute​.

To Find : Kinetic Energy of rotation of a circular disc .

Solution :

Mass m = 1 kg

Radius r = 0.2 m.

Disc makes 30/π rotations per minute​.

Here from Rotation of circular motion we know

$\[v = r\omega \]$                                         ( $\[\omega \]$ is angular velocity)

$\[v = r \times 2\pi n\]$

where,

n = no. of rotation per second

As disc makes 30/π rotations per minute​

Thus in Rotation per second,

n = $\[\frac{{30}}{\pi } \times \frac{1}{{60}}\]$              

Hence putting the value of n in original rotational value equation of v

$\[v = r \times 2\pi \times \frac{{30}}{\pi } \times \frac{1}{{60}}\]$

$\[v = r = 0.2\]$ m/s.

v = 0.2 m/s.

Kinetic Energy of Circular Disc is

$\[K.E. = \frac{1}{2}m{v^2}\]$                                ( Given : m = 1kg , v = 0.2 m/s)

$\[K.E. = \frac{1}{2} \times 1 \times {(0.2)^2}\]$

$\[K.E. = 0.02\]$ J.

Hence, Kinetic Energy of rotation of a circular disc is 0.02 J.