Physics, asked by SuzainShamim13, 11 months ago

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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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Answers

Answered by ʙʀᴀɪɴʟʏᴡɪᴛᴄh
3

Answer:

Hello

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 HeAvEnPrlnCesS
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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