- Rotational Motion
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- Angular velocity
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- 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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- Parallel Axis Theorem|Theorems of Moment of Inertia
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- Perpendicular Axis Theorem
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- Torque
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- work and power in rotational motion
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- Angular acceleration
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- Relationship between Angular momentum and torque
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- Conservation of Angular momentum
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- Radius of gyration
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- Rolling Motion|Kinetic Energy of rolling bodies
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- Rotational Motion problems with solutions

- 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

- The theorems are Parallel Axis theorem and Perpendicular axis theorem

- 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)

a. 7ML

b.7ML

c. 11ML

d.ML

Using parallel axis theorem

I=I

So x=L/2-L/4=L/4 Also I

So I=ML

Class 11 Maths Class 11 Physics Class 11 Chemistry

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