- 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

- We already know that the moment of inertia of a system about axis of rotation is given as

where m_{i}is the mass of the ith particle and r_{i}is its perpendicular distance from the axis of rotation

- For a system consisting of collection of discrete particles ,above equation can be used directly for calculating the moment of inertia

- For continuous bodies ,moment of inertia about a given line can be obtained using integration technique

- For this imagine dividing entire volume of the rigid body into small volume elements dV so that all the points in a particular volume element are approximately at same distance from the axis of rotation and le r be this distance

- if dm is the mass of this volume element dV,then moment of inertia may be given by

- Since density ρ of the element is defined as mass per unit volume so ρ=dm/dV hence equation (13) may be written as

- For homogeneous body density ρ is uniform hence ρ can be taken out of the integral sign i.e.

above integration can be carried out easily for bodies having regular shapes as can be seen from examples given below

- Consider a homogeneous and uniform rod of mass M and length L as shown below in the figure

- we have to calculate the moment of inertia of the rod about the bisector AB

- Consider middles point O to be the origin of the rod .Also consider an element of the rod between the distance x and x+dx from the origin. Since the rod is uniform so its density

ρ=M/L

Hence mass of the element dm=(M/L)dx - Perpendicular distance of this element from line AB is x,so that moment of inertia of this element about AB is

- For x=-L/2,the element is at the left end of the rod ,As x changes from -L/2 to +L/2 ,the element covers the whole rod

- Thus the moment of inertia of the entire rod about AB is

- Consider a uniform circular plate of mass M and radius R as shown below in the figure

- Let O be the center of the plate and OX is the axis perpendicular to the plane of the paper

- To find the moment of inertia of the plate about the axis OX draw two concentric circles of radii x and x+dx having these centers at O,so that they form a ring

- Area of this ring is equal to its circumference multiplied by its width i.e.

Area of the ring =2πxdx

- Mass of the ring would be

- Moment of inertia of this ring about axis OX would be

- Since whole disc can be supposed to be made up of such like concentric rings of radii ranging from O to R ,we can find moment of inertia I of the disc by integrating moment of inertia of the ring for the limits x=0 and x=R

- Consider a sphere of mass M and radius R .Let us divide this sphere into thin discs as shown in the figure

- If r is the distance of the disc then

- Volume of the disc would be

and its mass would be

dm=ρdV - Moment of inertia of this disc would be

Moment of inertia of the whole sphere would be

Factor 2 appears because of symmetry considerations as the right hemisphere has same MI as that of left one - Integration can be carried out easily by expanding (R
^{2}-x^{2})^{2}.On integrating above equation we find

Now mass of the sphere is

Hence

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