The rigid object shown is rotated about an
axis perpendicular to the paper and through
center point O. The total kinetic energy of
the object as it rotates is 8 J.

a) What is the moment of inertia of the ob-
ject? Neglect the mass of the connecting rods
and treat the masses as point masses.
Answer in units of kg · m2

b)What is the angular velocity of the object?
Answer in units of rad/s

Since the object is rotating, it will have a Rotational Kinetic Energy and Angular Velocity, that can be related by the equation below:

K.E= ½Iω2.

Defined as

Rotational kinetic energy = ½ moment of inertia * (angular speed)2..

K.E = 8J.

Therefore, making the moment of inertia, I the subject of the relation, we have

2*K.E = I * (Angular Velocity)2

Divide both sides by (Angular Velocity)2 and putting K.E = 8J,

(a) I = (2 x 8)/(Angular Velocity)2

I = 16/(ω2) Kg.m2

(b) Angular Velocity ω is calculated by making ω the subject of the relation

ω2 = (2 *K.E)/I

ω2 = (2 x 8)I = 16/I

Taking square root of both sides

ω = Sqrt(16/I) = 4/Sqrt(I) rad/s

Explanation:

When an object is rotating about its center of mass, its rotational kinetic energy is K = ½Iω2.

Rotational kinetic energy = ½ moment of inertia * (angular speed)2.

When the angular velocity of a spinning wheel doubles, its kinetic energy increases by a factor of four.

When an object has translational as well as rotational motion, we can look at the motion of the center of mass and the motion about the center of mass separately. The total kinetic energy is the sum of the translational kinetic energy of the center of mass (CM) and the rotational kinetic energy about the CM, Which is center O in the question.

d is the correct answer

nah.   i'm just sleepy.