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, θ₀ is initial angular displacement and ω₀ 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 F₁ be a force that is lying in planes perpendicular to the z-axis. The particle at P₁ describes a circular path of radius r₁ with center C on the axis: CP₁ = r₁

The point moves to position P₁' in time Δt

So, displacement ds₁ = r₁ d θ. The work done by force on particle is as follows:

dW₁ = F₁.d₁

= F₁ ds₁ cos φ

= F₁ (r₁ d θ)sin α₁

Where, φ is the angle between F₁ and the tangent at P₁ and α₁ is the angle between F₁ and the radius vector OP₁.

φ₁ + α₁ = 90°

Torque due to F₁ about the origin = OP₁ × F₁

OP₁ = OC + OP₁

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

So, effective torque τ₁ = CP × F₁

Magnitude of torque is τ₁ = r₁F₁ 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 P₀, the linear velocity V_r is directed exactly opposite to translational velocity V_cm. Since P₀ 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 P₁ at the top of the disc (V₁) 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))`