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Q.Derive an expression for the rotational Kinetic energy of a body and hence define moment of inertia of a body.

Himachal HpboseHPBOSE Himachal Pradesh Class 11 Board Exam 2024Subjective· 3mImportance★★★★★
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Summing (1/2)mivi^2 for every particle of a rotating rigid body and using vi = riomega gives K = (1/2)Iomega^2, where I = sum(miri^2) is defined as the moment of inertia.

Setup: Consider a rigid body rotating about a fixed axis with a constant angular velocity omega. Think of the body as made up of many tiny particles of masses m1, m2, m3, ... located at perpendicular distances r1, r2, r3, ... from the axis of rotation.

Since the body is rigid and rotates as a whole, every particle has the SAME angular velocity omega, but each particle's LINEAR speed depends on its own distance from the axis:

vi = ri * omega (for the i-th particle)

Kinetic energy of the whole body:

The kinetic energy of the i-th particle is (1/2)mivi^2 = (1/2)mi(riomega)^2 = (1/2)miri^2omega^2

Summing over all particles in the body, the total rotational kinetic energy is:

K = sum[(1/2)miri^2*omega^2] = (1/2)omega^2 * sum(miri^2)

Defining moment of inertia:

The quantity sum(mi*ri^2) depends only on the masses of the particles and how they are distributed relative to the axis (not on omega). This quantity is defined as the moment of inertia, I, of the body about that axis:

I = sum(mi*ri^2)

Substituting back:

K = (1/2) * I * omega^2

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