Rigid Body
* A rigid body is a ‘system of particles’ with a finite size that do not get deformed under the influence of any external force.
* The distance between ant two particles of an ideal rigid body does not change even when it is under the influence of external forces.
* The motion of a rigid body, in which the velocity of all the constituent particles is the same at any instant of time, is known as ‘Translational Motion’.
* This fixed line about which a rigid body rotates is called the ‘Axis of Rotation’.
* When a rigid body is in rotational motion, every point on it moves along a circular path with its centre on the axis of rotation and the plane of this circle perpendicular to the axis of rotation.
* The motion of any rolling body is a combination of rotational and translational motions.
* If a rigid body is fixed to an axis or point, it will have pure rotational motion.
* If the rigid body is free, it can have pure translational motion or a combination of translational and rotational motions.
Centre of mass
* The centre of a mass of system of particles is a point in space, whose location depends on the locations of constituent particles and their masses.
* When the masses of both particles are equal, x1 is a simple average of x1 and x2.
* We can regard the centre of mass of a system of particles as the mass weighted average location of the constituent particles.
* We can make use of symmetry considerations while finding the centre of mass of homogenous and regular shaped bodies.
* Based on similar symmetry considerations, we can prove that the centre mass of a thin rod, uniform disc, uniform sphere and other such regular bodies lie at their geometric centre.
Motion of centre of mass
* MA = Σmiai = Fext
* The motion of the centre of mass of a system of particles is independent of the internal forces.
* The centre of mass moves as if the whole mass of the system is concentrated at it and the external force is acting on it.
* The total momentum of the system of particles is equal to the product of “total mass of the system” and the “velocity of the centre of mass”.
* Fext = dP/dt. This is Newton’s second law of motion applicable for the system of particles.
* If the external force acting on the system of particles is zero, then
* Fext = 0 dP/dt = 0, P= constant
* This is law of conservation of momentum applicable to a system of particles.
* “if the total external force acting on a system of particles is zero, then the linear momentum of the system is constant”.
* Some times it is convenient to work with a reference frame that is attached to the centre of mass of the system of particles.
Vector product
* If the product of two vectors a and b, is a vector c, then c is called the vector product or cross product of a and b.
* The magnitude of a vector product is given by | C | = |a | |b| sin⍬,
* Where ⍬ = angle between vector a and b.
* Direction of a vector product can be determined using:
• Right handed screw rule
• Right hand rule
Properties of vector products:
• Not commutative
• Distributive with respect to vector addition
• Does not change its sign under reflection
Angular velocity and Angular Acceleration
* When the rigid body is in pure rotational motion, the angular displacement of all the particles of rigid body is the same and is the same and is equal to the angle through which the rigid body has rotated about the axis of rotation.
* In the SI system of units, angular displacement is measured in “Radian”.
* The ratio of the ‘Angular Displacement’ and ‘Time’ is defined as “Average Angular Velocity” of the particle.
*Angular velocity is defined as rate of angular displacement, ω, and is measured in “Radians per second”
* If the right hand fingers are curled in the direction of the rotation of the body, the direction of the stretched thumb is the direction of the angular velocity “ω”.
* The rate of change of angular velocity is defined as “Angular Acceleration”.
* We can represent relationship between linear velocity, and angular velocity, ω, as v= ω x r.
Torque and angular momentum
* The turning effect of a force is called torque or moment of force.
* Torque depends on the applied force, where it is applied and how it is applied.
* If a force ‘F’ is acting on a particle located at ‘r’, then the torque acting on the particle about the origin is a vector defined as r x F and is denoted by the Greek alphabet Ʈ(tau).
* Torque is a vector and its magnitude is expressed as Ʈ = r F sin⍬
* Where ⍬ is the angle between ‘r’ and ‘F’.
* The SI unit of torque is Nm.
* If ‘r’ represents the position vector of a particle of mass ‘m’ and linear momentum ‘p’, then the “angular momentum” of the particle about the origin is defined as the cross product of ‘r’ and ‘p’.
* The magnitude of angular momentum of the particle is ⍳ = r p sin⍬.
* The SI unit angular momentum is kg m2s-1
* We can find a useful relationship between “torque” and “angular momentum”. We have, ⍳ =r Xp.
* The rate of change of angular momentum of a particle with time is equal to the torque acting on the particle.
Equilibrium of a rigid body
* If the net force acting on a body is zero, it moves with constant momentum. In such a case we say the body is in “translational equilibrium”.
* If the vector sum of all the torques acting on a rigid body is equal to zero, it is said to be in “rotational equilibrium”.
* A rigid body is in mechanical equilibrium when the net external force and net external torque acting on it is zero.
* If a body in mechanical equilibrium is at rest, then the body is said to be in “static equilibrium”.
* Two parallel forces that are equal in magnitude and opposite in direction, acting on a rigid body with different lines of action forms a “couple”.
* Moment of couple Ʈ = r X F, where ‘r’ is the perpendicular distance between two forces.
* A couple is needed to make a rigid body rotate without translational motion.
* For a lever, effort X effort arm = load X load arm. This statement is known as “the
* principle of moments for a lever”.
* The mechanical advantage of a lever =effort arm/ load arm.
Centre of gravity
* The centre of gravity of a rigid body is defined as that point where the total gravitational torque on the body is zero.
* The weight of a rigid body can’t produce any net torque, when it is calculated about the centre of gravity.
* Torque due to the weight of a body can be calculated by assuming the whole weight of the body concentrated at its centre of gravity.
* If we want to balance a rigid body, we need to apply a force in the upward direction that also passes through the centre of gravity of the body.
* The two concepts “centre of mass” and “centre of gravity” are two different concepts.
* If the acceleration due to gravity is the same for all particles of a rigid body, the “centre of gravity” coincides with its “centre of mass”.
Moment of inertia
* The moment of inertia is defined as, І= Σmiri2
* The rotational kinetic energy of a rigid body is k = ½ Іω2
* ‘І ‘is the rotational analogue of mass ‘m’ in linear motion.
* ‘І ‘is called the moment of inertia of a rigid body.
* Moment of inertia is the measure of resistance offered by a body to any change in its rotational motion.
* Moment of inertia is also known as rotational inertia.
* Moment of inertia of a rigid body depends on the mass of the body, distribution of mass about the axis of rotation and the position and orientation of the axis of rotation.
* The SI unit of moment of inertia is kg m2 and its dimensional formula is (M1L2)
* For a continuous distribution of mass, the moment of inertia is expressed as І= ∫r2dm
* Fly wheel is a mechanical device with a high moment of inertia and so it resists any change in its rotational motion.
Moment of inertia of regular shaped bodies
* The moment of inertia of a system of particles is given by І =Σmiri2
* For an isolated body of mass ‘m’ and at a distance ‘r’ from the axis of rotation, І=mr2
* The moment of inertia of inertia of a thin uniform ring of mass ‘m’ and radius ‘r’, about an axis perpendicular to the plane of the ring and passing through its centre is “mr2”
* The moment of inertia of hollow cylinder of mass ‘m’ and radius ‘r’ about the axis of cylinder is also equal to “mr2”.
* The moment of inertia of a thin rod of mass ‘m’ and length ‘l’, about an axis perpendicular to the rod and passing through the midpoint is given by the expressions, I=ml2/12
* The moment of inertia of a solid sphere of mass ‘m’ and radius ‘r’ about an axis passing
through its centre is I=2mr2/5.
Theorem of perpendicular and parallel axis
* The “Perpendicular Axis Theorem” is applicable only to plane bodies.
* The “Perpendicular Axis Theorem” states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of its moments of inertia about two perpendicular axis lying in the plane of the lamina and intersecting each other at a point where the axis perpendicular to the plane of the lamina passes through.
* The “Parallel Axis Theorem” states that “the moment of inertia of a rigid body about any axes, is equal to the moment of inertia about a parallel axes passing through its centre of mass plus the product of the mass of the body and the square of the distance between the two parallel axes”.
Kinematics of rotational motion about a fixed axis
* The rotational analogy to “linear motion of translational motion” is “rotational of a rigid body about a fixed axis”.
* Corresponding to “linear displacement” in translational motion, we have “angular displacement” in rotational motion.
* Angular velocity, ω, in rotational motion plays the same role as “linear velocity”,v, in translational motion.
* Since the axis of rotation is fixed, the direction of angular velocity is always along the axis of rotation and hence it need not be treated as a vector.
* “Angular acceleration” is the rotational analogy of “linear acceleration” of translational motion.
* The following are the kinematic equations for rotational motion with uniform angular acceleration:
• ω = ω0 +ᾳt
• ⍬ = ⍬0 +ω0t +1/2ᾳt2
• ω2=ω02+2a (⍬-⍬0)
Dynamics of rotational motion about a fixed axis
* The work done by the total torque acting on a body rotating about a fixed axis is dw =Ʈd⍬.
* We know that, the rate of doing work is known as power.
* Instantaneous power, P=dW/dt = Ʈω
* When work is done on rotating body, its energy must increase.
* Therefore, the rate of work done on the rigid body rotating about a fixed axis must be equal to the rate of increase in its rotational kinetic energy.
* We know that, the rotational kinetic energy of a rigid body is denoted by Kr = ½ Iω2.
* The equation Ʈ = Iᾳ is the Newton’s Second Law for rotational motion.
* It states that, the angular acceleration of a rotating body is directly proportional to the applied “torque” and is inversely proportional to the “moment of inertia” of the body.
Angular momentum and conservation of angular momentum
* The angular momentum of a system of particles is denoted by L= Σri X pi.
* In general for a rigid body rotating about a fixed axis, its angular momentum “L” does not lie along the axis of rotation.
* If the external torque is zero, then dLz/dt = 0 or d(Iω)/dt = 0
* Hence, LZ = Iω = constant.
* The law of conservation of angular momentum states that in the absence of net external torque, the total angular momentum of the system remains conserved.
Rolling motion
* A rolling body undergoes simultaneous rotational and translational motion.
* Friction is required to start rolling.
* Rolling without slipping is known as “pure rolling”.
* When the wheels roll without slipping, at any instant of time, the point of the wheel in contact with the surface is instantaneously at rest with respect to the surface.
* No static frictional force acts on the wheels when they are in pure rotation.
* Condition for pure rolling: VCM = ωR.
* When a body is in pure rotation, the velocity of the centre of mass VCM = Ωr and the velocity of the point of contact, P0, VP0 =0, and the velocity of a point at the top, P1, VP1 =2 ωR.
* An axis passing through the instantaneous point of contact “P0” and along the angular velocity “ω” that is perpendicular to the plane of the disc is called the “instantaneous axis of rotation”.
* Rolling without slipping is not possible on a smooth inclined surface.
* While solving problems relating to the rolling of an object on an inclined surface, we can use two different methods.
a) In the first method, we analyse the forces and torques acting on the object.
b) The second method involves the use of the Law of Conservation of energy.
* For an object in pure rotation, the work done by the frictional force is zero.
KE = KCM +Kr
KE = ½ MVCM2 +1/2 Iω2