 # Dynamics of Rotational Motion about a Fixed Axis We already know that a rigid body may follow a translational or rotational motion. There may be objects which follow both, hence while studying the rigid body dynamics of such objects we compare the kinematics of both the motions. The page below deals with the rigid body dynamics that follows rotational motion about a fixed axis.

### Suggested Videos        ## Rigid Body Dynamics of Rotational Motion

When in motion, a rigid body is believed to be a system of particles. Each of its particles follows a path depending on the kind of motion it follows. In a translational motion, all the particles move and behave in a similar manner.  But in rotational motion, the rigid body dynamics indicate a different behaviour.

All the particles behave differently. Since rotation here is about a fixed axis, every particle constituting the rigid body behaves to be rotating around a fixed axis. As the distance from the axis increases the velocity of the particle increases. A particle in rotational motion moves with an angular velocity. Moment of inertia and torque for the rotational motion are like mass and force in translational motion. These analogues for both the motions give us the idea of a particle’s behaviour while in motion. While describing the rigid body dynamics during rotational motion about a fixed axis we intend to highlight the relation between the analogues of both the motions.

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Let us learn in detail about Rotational Dynamic.

## Components of Interest in Rotational Motion

In the rotational motion of a particle about a fixed axis, we take into consideration only those components of torques that are along the direction of the fixed axis. Since these are the components that cause the body to rotate, it is necessary that every component of torque be to lie in a plane perpendicular to the axis of rotation.

The axis will turn from its position if the component of torque is perpendicular to the axis. The fixed axis of rotation is maintained only when the external torque is constrained by necessary forces. This is the reason we do not consider the perpendicular component of torques. We, therefore consider only the forces that lie in the plane perpendicular to the axis.

All the forces that are parallel to the rotational axis will apply torque perpendicular to the axis, hence should be ignored. Another thing to be considered are those components of position vectors that are perpendicular to the axis. Any component of position vector that is along the axis will give a torque perpendicular to the axis and hence can be ignored.

## Graphical Representation of a Rigid Body Dynamics

For understanding the dynamics of a rigid body during rotational motion around a fixed axis we need to study the graph below: The graph above represents work done by a torque acting on a particle that is rotating about a fixed axis. The particle is moving in a circular path with center Q on the axis. P1P’1 is the arc of displacement ds1. The graph in the figure shows the rotational motion of a rigid body across a fixed axis.

As already mentioned that while studying rigid body dynamics of rotational motion we consider only those forces that lie in planes perpendicular to the axis of rotation. F1 is the same force acting on particle P1 and lies in a plane perpendicular to the axis.This plane can be called as x’-y’ plane and r1 is the radius of the circular path followed by particle P1.

Now from the figure, we know that QP1 = r1 and particle P1 moves to position P’ in time Δt. The displacement of the particle here is ds1. The magnitude of ds1 = r1dθ. Here, dθ is the angular displacement of the particle and is equal to ∠P1QP’1. The work done by the force on particle = dW1 =  F1ds1cosø= F1(r1dθ) sinα1.

ø1 is the angle between the tangent at P1 and the force F1 and αis the angle between radius vector and F1.

## Torque

In the figure, we notice that the radius vector is 90° (ø1 + α1). The torque due to force F1 around the origin is a product of radius vector and F1. Here we should remember that any particle on the axis is excluded from our analysis. So the effective torque in that case arising due to force F1 is signified by τ.

τ= QP × F1. Effective torque τ is directed along the axis of rotation with a magnitude of r1F1sinα. This brings us to the conclusion that work done = τ1 dθ.

## Workdone by Multiple Forces

The above graph represents the force and work done by the same force on one particle. Now let’s consider the case of more than one forces acting on the rigid body. In a system of particles with more than one force acting on the system, the work done by each of them is added, this gives the total work done by the body. The magnitude of torques due to various forces is denoted as τ1, τ2,τ3…… etc. So, dW = (τ1+ τ2+τ3 +……) dθ.

The angular displacement (dθ) for all the particles is same here. All the torques under our consideration are parallel to the fixed axis and the magnitude of the total external force is just the sum of individual torques by various particles. This gives us the equation: dW = τ dθ. This represents the work done by the total torque that acts on the rigid body rotating about a fixed axis.

## Relation Between Rotational and Translational Motion

The workdone (dW) by external torque during rotational motion = τ dθ. The workdone (dW) by the external force during translational motion = Fds. Here, ds is the linear displacement and F is the external force. Dividing both by dt we get: P= dW/dt = τ dθ/dt = τω or P= τω. Here, P is the instantaneous power. Instantaneous power (P) for linear motion = Fv.

Do you notice any similarity? We know that a perfectly rigid body lacks internal motion hence the work done by external torque increases the kinetic energy of the body. Now, to find the rate of increase in kinetic energy we equate the equations as: d/dt [Iω2]/ 2 = I (2ω)/2 [dω/dt]. From the equation, we assume that moment of inertia is constant and does not change with time.

This also means that the mass of the body also does not change. Now since the axis is also fixed, its position with respect to the body also doesn’t change so: using α= dω /dt. d/dt [Iω2]/ 2 = Iωα. Therefore the rate of work done and rate of increase in kinetic energy can be equated as τω= Iωα or τ= Iα. Now, this equation is similar to Newton’s Second law for Linear motion, F = ma.

Therefore from the relation of work done and kinetic energy we come to a conclusion that Newton’s second law of linear motion is applicable to rigid bodies undergoing rotational motion. So, Newton’s second law of rotational motion states that the angular acceleration during rotational motion of a rigid body is directly proportional to the applied torque and inversely proportional to the moment of Inertia of that body.

## Sample Question For You

Q: A train engine that weighs 5000N stops at the exact center of a bridge of weighs 75,000N and has two equally spaced pillars to support the bridge. Find out the sum of the torques?

A) 75000 N     B) 40000 N      C) 2500 N      D) 0 N

Solution:D) The bridge + train engine system is in equilibrium. Therefore, the net torque about every point must be zero. Hence the correct option is 0 N.

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