- 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

- Angular acceleration is the rate of change of angular velocity with respect to time. Thus for rigid body rotating about a fixed axis

- Unit of angular acceleration is radian /sec
^{2}and it is a vector quantity like Angular velocity

- Now Let's derive it for rotation around fixed axis
- Consider a rigid body of arbitrary shape rotating about a fixed axis through point O and perpendicular to the plane of the paper as shown below in the figure-1

- while the body is rotating each and every point in the body moves in a circle with their center lying on the axis of rotation and every point moves through the same angle during a particular interval of time

- Consider the position of a particle say i
^{th}particle at point P at a distance r_{i}from point O and at an angle θ_{i}which OP makes with some reference line fixed in space say OX as shown below in the figure

- If particle moves an small distance ds
_{i}along the arc of the circle in small amount of time dt then

ds_{i}=v_{i}dt ----(1)

where v_{i}is the speed of the particle - dθ is the angle subtended by an arc of length ds
_{i}on the circumference of a circle of radius r_{i},so dθ( in radians) would be equal to the length of the arc divided by the radius

i.e.

dθ=ds_{i}/r_{i}=v_{i}dt/r_{i}----(2) - distance ds
_{i}would vary from particle to particle but angle dθ swept out in a given time remain same for all the particles i.e. if particle at point P moves through complete circle such that

dθ=2π rad

Then all the other particles of the rigid body moves through the angular displacement dθ=2π

- Now We already know about angular velocity is given by ω=dθ/dt ----(3)
- Putting equation (3) in equation (2) we find

v_{i}=r_{i}(dθ/dt) =r_{i}ω ---(4)

This shows that velocity of ith particle of the rigid body is related to its radius and the angular velocity of the rigid body

- Now angular acceleration can be written as

$\alpha = \frac {d \omega}{dt}$

Now from equation 3

$\alpha = \frac {d}{dt} \frac {d \theta}{dt} = \frac {d^2 \theta}{dt}$

- Angular acceleration holds not only for that rotating rigid body but also for the each and every particle of that body

- Differentiating equation (4) w.r.t to t we find

where a_{i}=a_{it}=r_{i}α is the tangential component of linear acceleration of a point at a distance r_{i}from the axis - The above equation is important as it shows the relation ship the tangential components of linear acceleration of a point and angular acceleration of a point
- Each particle in the rigid body also has the radial linear acceleration component v
^{2}/r ,which can also be expressed in terms of an of angular velocity i.e.

and this acceleration a_{c}pointing inwards towards the radial line is also known as centripetal acceleration - So total accleration of an point in the rotation can be expressed as.

$a = \sqrt {a_c^2 + a_i^2}$

a. increases in magnitude but retains the same angle with the tangent to the rim

b.increases in magnitude and becomes more nearly radial

c. increases in magnitude and becomes more nearly tangent to the rim

d. decreases in magnitude and becomes more nearly radial

Tangential acceleration=radius* angular acceleration

Since angular acceleration is constant,Tangential acceleration is constant

Radial acceleration=r* (angular velocity)

Since angular velocity increase with time,Radial acceleration increase with time

So resultant acceleration increase with time and becomes more radial as time passes

Two wheels are identical but wheel B is spinning with twice the angular speed of wheel A. The ratio of the magnitude of the &radical acceleration of a point on the rim of B to the magnitude of the radial acceleration of a point on the rim of A is:

a. 4

b. 2

c. 1/2

d. 1/4

Radial acceleration=r* (ω)

For wheel A

Radial acceleration of A =r* (ω)

For wheel B

Radial acceleration of B=r* (2ω)

So Radial acceleration of B/Radial acceleration of A=4:1

a 2

b 1/2

c 4

d 1/4

At rim

$v=r \omega$

At point between the center and rim

$v=\frac {r}{2)} \omega $

Ratio =2

Class 11 Maths Class 11 Physics Class 11 Chemistry