# Rotational Motion

## Kinematics of Rotational Motion about a Fixed Axis

You are aware about the kinematic equations of linear motion. Corresponding equations for rotational motion with uniform angular acceleration are as follows:

`ω=ω_0+αt`

`θ=θ_0+ω_0t+1/2αt^2`

`ω^2=ω^2+2α(θ-θ_0)`

Where, θ_{0} is initial angular displacement and ω_{0} is initial angular velocity of the rotating body.

### Dynamics of Rotational Motion

We need to consider only those components of torques which are along the direction of the fixed axis, because axis is fixed. A component of the torque perpendicular to the axis will tend to turn the axis from its position. So, perpendiclar components of torque need not be taken into account. This means the following:

- We need to consider only those forces which lie in planes perpendicular to the axis. A force which is parallel to the axis will give torques perpendicular to the axis and so it does not need to be taken into account.
- Only those components of position vectors are cosidered which are perpendicular ot the axis. Components of position vectors along the axis will produce torques perpendicular to the axis.

## Work done by Torque

This figure shows rotation of a rigid body about a fixed axis, i.e. z-axis. Let F_{1} be a force that is lying in planes perpendicular to the z-axis. The particle at P_{1} describes a circular path of radius r_{1} with center C on the axis: CP_{1} = r_{1}

The point moves to position P_{1}' in time Δt

So, displacement ds_{1} = r_{1} d θ. The work done by force on particle is as follows:

dW_{1} = F_{1}.d_{1}

= F_{1} ds_{1} cos φ

= F_{1} (r_{1} d θ)sin α_{1}

Where, φ is the angle between F_{1} and the tangent at P_{1} and α_{1} is the angle between F_{1} and the radius vector OP_{1}.

φ_{1} + α_{1} = 90°

Torque due to F_{1} about the origin = OP_{1} × F_{1}

OP_{1} = OC + OP_{1}

As OC is along the axis, let us exclude the torque resulting from it.

So, effective torque τ_{1} = CP × F_{1}

Magnitude of torque is τ_{1} = r_{1}F_{1} sin α

So, `dW_1=τ_1dθ`

Or, `dW=τdθ`

This expression gives the work done by the total (external) torque τ which acts on the body about a fixed axis.

Now, instantaneous power can be given as follows:

`P=(dW)/(dt)`

`=τ(dθ)/(dt)=τω`

Or, `P=τω`

## Rolling Motion

Let us take a disc which is rolling without slipping. This means that at any instance of time, the bottom of the disc (in contact with the surface) is at rest on the surface.

Let us assume that velocity of center of mass = V_{cm}

This is the translational velocity of the disc. Translational motion is parallel to the level surface.

Velocity at any point of the disc has two parts, translational velocity V_{cm} and linear velocity V_{r}.

Magnitude of **V _{r}** = V

_{r}= rω

Where, r = distance of particle from center.

At P_{0}, the linear velocity V_{r} is directed exactly opposite to translational velocity V_{cm}. Since P_{0} is instantaneously at rest, hence V_{cm} = Rω.

So, for the disc the condition for rolling without slipping is

`V_(cm)=Rω`

This means that the velocity of point P_{1} at the top of the disc (**V _{1}**) has a magnitude as follows:

`v_(cm)+Rω=2v_(cm)`

#### Kinetic Energy of Rolling Motion

Kinetic energy of rolling body can be given by following equation:

`K=K'+(MV^2)/2`

Where, K' is the kinetic energy of rotational motion and `(MV^2)/2` is the kinetic energy or translational motion.

Kinetic energy of rolling motion can be written as follows:

`K'=(Iω^2)/2`

Where, I is the moment of inertia about the appropriate axis.

So, kinetic energy of rolling body can be given as follows:

`K=1/2Iω^2+1/2mv_(cm)^2`

We know, `I=mk^2` where k is the corresponding radius of gyration.

We also know v_{cm} = R ω

Substituting these values in above equation, we get

`K=1/2(mk^2v_(cm)^2)/(R^2)+1/2mv_(cm)^2`

Or, `K=1/2mv_(cm)^2(1+(k^2)/(K^2))`