- Introduction
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- Angular velocity and angular acceleration
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- Rotation with constant angular acceleration
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- Kinetic energy of Rotation
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- Calculation of moment of inertia
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- Theorems of Moment of Inertia
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- Torque
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- work and power in rotational motion
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- Torque and angular acceleration
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- Angular momentum and torque as vector product
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- Angular momentum and torque of the system of particles
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- Angular momentum of the system of particles with respect to the center of mass of the system
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- Radius of gyration
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- Kinetic Energy of rolling bodies (rotation and translation combined)
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- Solved examples

- There are two general theorems which proved themselves to be of great importance on moment of inertia

- These enable us to determine moment of inertia of a body about an axis if moment of inertia of body about some other axis is known

- This theorem is applicable only to the plane laminar bodies

- This theorem states that, the moment of inertia of a plane laminar about an axis perpendicular to its plane is equal to the sum of the moment of inertia of the lamina about two axis mutually perpendicular to each other in its plane and intersecting each other at the point where perpendicular axis passes through it

- Consider plane laminar body of arbitrary shape lying in the x-y plane as shown below in the figure

- The moment of inertia about the z-axis equals to the sum of the moments of inertia about the x-axis and y axis

- To prove it consider the moment of inertia about x-axis

where sum is taken over all the element of the mass m_{i}

- The moment of inertia about the y axis is

- Moment of inertia about z axis is

where r_{i}is perpendicular distance of particle at point P from the OZ axis

- For each element

r_{i}^{2}=x_{i}^{2}+ y_{i}^{2}

- This theorem relates the moment of inertia about an axis through the center of mass of a body about a second parallel axis

- Let I
_{cm}be the moment of inertia about an axis through center of mass of the body and I be that about a parallel axis at a distance r from C as shown below in the figure

Then according to parallel axis theorem

I=I_{cm}+Mr^{2}where M is the total mass of the body

- Consider a point P of the body of mass m
_{i}at a distance x_{i}from O

- From point P drop a perpendicular PQ on to the OC and join PC.So that

OP^{2}=CP^{2}+ OC^{2}+ 2OC.CQ ( From geometry) and m_{i}OP^{2}=m_{i}CP^{2}+ m_{i}OC^{2}+ 2m_{i}OC.CQ

- Since the body always balances about an axis passing through center of mass, so algebraic sum of the moment of the weight of individual particles about center of mass must be zero. Here

which is the algebraic sum of such moments about C and therefore eq as g is constant

Thus we have I=I_{cm}+ Mr^{2}---(17)

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