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Question:- A solid cylinder of mass 20kg rotates about its axis with angular speed 100rad/s ,the radius of the cylinder is 0.25m . What is the magnitude of the angular momentum of the cylinder about its axis.
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Answered by
3
Answer:
Explanation:
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
Answered by
3
Answer:
Given,
Mass of cylinder (M) = 20 Kg
Radius of its ( R) = 0.25 m
Angular speed (w ) = 100 rad/s
We know, moment of inertia of solid cylinder ( I) = (1/2)MR²
= (1/2) × 20× (0.25)²
= 0.625 kg.m²
Kinetic energy associated with the rotation of the cylinder is given by
K.E = (1/2)Iw²
= (1/2) × 0.625 × (100)²
= 3125 joule
angular momentum = Iw = 0.625* 100 = 62.5 js
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