Question

A disc of diameter $0.4m$and of mass $10kg$is rotating about its axis at the rate of $1200rev/\min $
Find
 (a) Angular momentum
 (b) Rotational kinetic energy

Answer 1

Hint:
As the direct value is not given we need to split the direct formula. And for the rotational kinetic energy we can think of the example of earth rotation.

Formula used:
$
  L = I \times \omega \\
   \Rightarrow L = (m{r^2}) \times (2\pi v) \\
 $
Where L is the angular momentum.
It is the rotational inertia.
ω is the angular velocity.
m is the Mass given here
R is the radius
V is the velocity
$
  {(K.E)_R} = \dfrac{1}{2}I{\omega ^2} \\
   \Rightarrow {(K.E)_R} = \dfrac{1}{2} \times \dfrac{{m{R^2}}}{2} \times {\omega ^2} \\
 $
Where ${(K.E)_R}$ is the rotational kinetic energy
And rest all quantities are same as described above equation

Complete step by step solution:
Let us first understand about angular momentum
Angular momentum is the property of a rotating body given by the product of the moment of inertia and the angular velocity of the rotating object.
Here given quantities to find the angular momentum is
 $d = 0.2m$Radius obtained is $r = \dfrac{d}{2} = \dfrac{{0.4}}{2} = 0.2m$
$m = 10kg$
Angular velocity $\omega = 1200rev/\min = 20rev/\sec $
So according to this case this object is rotating about a fixed point.
So here the formula is used is
$
  L = I \times \omega \\
   \Rightarrow L = (m{r^2}) \times (2\pi v) \\
    \\
 $
$
  L = I \times \omega \\
   \Rightarrow L = (m{r^2}) \times (2\pi v) \\
   \Rightarrow L = 10 \times {(0.2)^2} \times 2 \times \dfrac{{22}}{7} \times 20 \\
   \Rightarrow L = 25.14kg{m^2}/s \\
 $
Rotational energy occurs due to the rotation of the object and is a part of its total kinetic energy. Rotational energy also known as angular kinetic energy.
Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity

$
  {(K.E)_R} = \dfrac{1}{2}I{\omega ^2} \\
   \Rightarrow {(K.E)_R} = \dfrac{1}{2} \times \dfrac{{m{R^2}}}{2} \times {\omega ^2} \\
   \Rightarrow {(K.E)_R} = \dfrac{1}{2} \times \dfrac{{10 \times 0.2 \times 0.2}}{2} \times \left( {\dfrac{{1200 \times 2\pi }}{{60}}} \right) \\
   \Rightarrow {(K.E)_R} = 1580J \\
 $

ADDITIONAL INFORMATION:
Let’s take an example
A ball that rolls down a ramp rotates as it travels downward. The ball has rotational kinetic energy from the rotation about its axis and translational kinetic energy from its translational motion
An ice-skater when she starts to spin she first starts off with her hands and legs far apart from the centre of her body. But when she needs more angular velocity to spin, she gets her hands and leg closer to her body. Hence, her angular momentum is conserved and she spins faster.

Note:
Mostly in the question all the required quantities will be given. But sometimes we need to split the quantities of the direct formula and we can easily find the solution