Question

A solid cylinder of mass 20kg rotates about its axis with an angular speed under the radius of the radius of the cylinder is 0.25 m what is the kinetic energy associated with the rotation of the cylinder what is the magnitude of the angular momentum of different about its axis

Answer 1

Answer:

62.5 j

Explanation:

Mass of cylinder (M) = 20 Kg (Given)

Radius of its ( R) = 0.25 m  (Given)

Taking the angular speed (w) as = 100 rad/s

Moment of inertia of the solid cylinder ( I) = (1/2)MR²

= (1/2) × 20× (0.25)²

= 0.625 kgm²

Kinetic energy associated with the rotation of the cylinder is given by equation -

K.E = (1/2)Iw²

= (1/2) × 0.625 × (100)²

= 3125 joule

Thus, the angular momentum = Iw = 0.625 × 100 = 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