Question

A stone of mass 4kg is whirled in a horizontal circle of radius 1m and makes 2rev./sec the moment of inertia of the stone about the axis of rotation is

Answer 1

The moment of inertia of the stone about the axis of rotation is 4 Kg/m^2.

Explanation:

Given data:

Mass of stone = 4 kg

Radius of circle "r" = 1

Number of revolutions = 2 rev./sec

solution:

Formula of moment of inertia about the axis is given by.

I = MR^2

I = 4 x 1^2

I = 4 Kg/m^2

Hence the moment of inertia of the stone about the axis of rotation is 4

Kg/m^2.

Also learn more

A flywheel of moment of inertia 2 kgm^2 is rotated at a speed of 30 rad/sec. A tangential force at the rim stops the wheel in 15 sec. Average torque of the force is ?

https://brainly.in/question/1091259

Answer 2

Answer: 4kgm^2

Explanation:

no use of 2rev/sec , we just have to find MOI

we need mass and radius.

I=MR^2

mass=4 r=1

I=4kg *1^2

=4kgm^2

hope this helps ..

have a nice day..