Question

A solid cylinder of mass 20kg rotates about its axis with angular speed 100 rad s⁻¹. The radius of the cylinder is 0.25 m. What is the kinetic energy associated associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Answer 1

first of all, we have to find Kinetic energy of the rotating cylinder.

We know that, Rotational kinetic energy,
KE = 1/2Iω²
WhereI is moment of inertia
ω is angular velocity
Given
Mass of cylinder, M = 20 kg
Radius of cylinder, R = 0.25 m = 1/4 m
Angular speed of cylinder, ω = 100 rad/s
Moment of inertia of a solid cylinder, I = 1/2 MR²
I.e. I = 1/2 × 20 × (1/4)²
= 5/8 kg m²

∴ KE = 1/2 I ω²
= 5/16 (100)²
= 3125 J
= 3.125 kJ

We need to find angular momentum of rotating cylinder about its axis.
Angular momentum, L = I ω
We know that,
Moment of inertia of cylinder, I = 5/8 kg m²
Angular speed of cylinder, ω =100 rad/s
∴ Angular momentum, L = 5/8 × 100
= 62.5 kg m/a
= 62.5 Js

Answer 2

Answer:

Mass of the cylinder, m = 20 kg

Angular speed, ω = 100 rad s–1

Radius of the cylinder, r = 0.25 m

The moment of inertia of the solid cylinder:

I = mr2 / 2

= (1/2) × 20 × (0.25)2

= 0.625 kg m2

∴ Kinetic energy = (1/2) I ω2

= (1/2) × 6.25 × (100)2 = 3125 J

∴Angular momentum, L = Iω

= 6.25 × 100

= 62.5 Js