Question

A solid cylinder of moment of interia 0.62 kg m2 rotates about its axis with angular speed 100 rad s-1 what is magnitude of angular momentum of cylinder about its axis
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Answer 1

Answer:

ANSWER

The moment of inertia of a solid cylinder=mr

2

/2

=

2

1

×20×(0.25)

2

=0.625kgm

2

Therefore kinetic energy=

2

1

Iω

2

=3125J

Angular momentum ,L=Iω

=0.625×100

=62.5Js

Answer 2

The magnitude of the angular momentum of the cylinder around its axis is 62 kg.m²/s

Given: A strong cylinder with a moment of inertia of 0.62 kg,m² rotates approximately its axis with an angular speed of 100 rad.s-1

To Find: The magnitude of the angular momentum of the cylinder about its axis.

Solution:

• The moment of inertia of a body may be described as an opposition that a body exhibits to having its speed of rotation about an axis altered by the application of a torque.

• The property of any rotating item given by the moment of inertia times the angular velocity is called the angular momentum of the frame.

• The connection between the moment of momentum and the moment of inertia may be given by way of the components,

                      L = I × ω                                                                 ...(1)

wherein L = moment of momentum, I = moment of inertia, ω = angular speed.

whilst we get to numerical we are given;

I = 0.62 kg.m², ω = 100 rad/s

Plugging the precise values into (1), we get;

                       L = I × ω

                   ⇒ L = 0.62 x 100

                   ⇒ L = 62 kg.m²/s

The magnitude of the angular momentum of the cylinder around its axis is therefore 62 kg.m²/s

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