- Introduction
- Angular velocity and angular acceleration
- |
- 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

a. L

b. L/4

c. L/2

d. 2L

Angular momentum for this is defined as

=mr

First case

L=mr

Second case

L

So L

a. is zero

b. remains constant

c. goes on increasing

d. goes on decreasing

L=(mv)Xr

or

L=mvrsinθ

Now rsinθ=perpendicular distance from x axis which is constant

So Angular momentum is constant

a. Up the incline while ascending and down the incline while descending

b.Up the incline while ascending and desending

c. down the incline while ascending and up the incline while descending

d.down the incline while ascending and desending

Imagine the cylinder to be moving on a frictionless surface.In both the cases the acceleration of the CM of the cylinder is gsinθ.This is also the acceleration of the point of contact

of the cylinder and the inclined plane..Also no torque (about the center of the cylinder) is acting on the cylinder since we assumed the surface to be a frictionless and the forces

acting on the cylinder is mg and N which passes through the center of the cylinder.Therefore the net movement of the point of contact in both the cases is in downward direction

Therefore frictional force will act in upward direction in both the cases

a. 28.6 m

b 30 m

c. 28 m

d. none of these

Let h be the height

The rotational and translational KE of the ball at the bottom will be changed to Gravitational energy when the sphere stops .

We therefore writes

(1/2)Mv

For a solid sphere I=(2/5)Mr

So (1/2)Mv

or

v

or h=28.6 m

a. (2/3)gsinθ

b.(2/3)gcosθ

c gsinθ

d none of these

Net force on the cylinder

F

or ma=mgsinθ -f

Where f is the frictional force

Now τ=fXR=Iα

Now in case of pure rolling we know that

a=αR => α=a/R

So f=Ia/R

From 1 and 2

a=mgsinθ /[m+(I/R

Now I=mR

So a=(2/3)gsinθ

a. 5.26 rev/s

b. 5.2 rev/s

c 4.7 rev/s

d. none of thes

Law of conservation of angular momentum

I

or

1.33(1.9)=.48ω

or

ω

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

acceleration vector for a point on the rim:

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 resulttant 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

the speed of a point on the rim to the speed of a point halfway between the center and the rim is:

a 2

b 1/2

c 4

d 1/4

At rim

v=rω

At point between the center and rim

v=(r/2)ω

Ratio =2

slowing at first. What time elapses before its angular velocity is18 rad/s in the direction opposite to its initial

angular velocity?

a 3 sec

b 6 sec

c 18 sec

d none of these

ω

ω=-18

anugular acceleration(α)=-2

Now

ω=ω

or t=18

(A) I

(B) I

(C) I

(D) I

Where D

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