Topics are arranged as per SYLLABI published by Maharashtra State Board of Secondary and Higher Secondary Education.
In rotation of a rigid body about a fixed axis, every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis.
The following table summarizes the rotational motion analogous of the basic concepts for linear or translatory motion.
Linear Motion | Rotational Motion |
---|---|
Distance/displacement (s)t | Angle or angular displacement |
Linear velocity | Angular velocity |
Linear acceleration | Angular acceleration |
Mass (m) | Moment on inertia (I) |
Linear momentum, p = mv | Angular momentum, L = Iω |
Force, F = ma | Torque, |
Also, Force | Also, torque, |
Translational KE, | Rotational KE, |
Work done, W = Fs | Work done, |
Power, P = Fv | Power, |
Linear momentum of a system is conserved when no external force acts on the system. | Angular momentum of a system is conserved when no external torque acts on the system. |
Equation of translator motion | Equation of rotational motion |
It is a quantity expressed by, a body’s tendency to resist angular acceleration, which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation.
Moment Of Inertia Of Some Object:
Body | Axis of Rotation | Moment of Inertia (I) |
---|---|---|
Uniform circular ring of mass M and radius R | About an axis passing through centre and perpendicular to its plane. | MR^{2} |
About an diameter | 0.5 MR^{2} | |
About an tangent in its own plane | 1.5 MR^{2} | |
About a tangent perpendicular to its plane | 2MR^{2} | |
Uniform Circular Disc of mass M and radius R. | About an axis passing through centre and perp. To its plane. | 0.5MR^{2} |
About an diameter | 0.25 MR^{2} | |
About an tangent in its own plane | 1.25 MR^{2} | |
About a tangent perpendicular to its plane | 1.5MR^{2} | |
Solid sphere of radius R and mass M. | About an diameter. | 0.4MR^{2} |
About a tangential axis | 1.4 MR^{2} | |
Spherical shell of radius R and mass M | About an diameter. | 0.67MR^{2} |
About a tangential axis | 1.67 MR^{2} | |
Long thin rod of Length L | About an axis through C.G. and perpendicular to rod. | |
About an axis through one end and perpendicular to rod. |
Consider a rigid object rotating about a fixed axis at a certain angular velocity. Since every particle in the object is moving, every particle has kinetic energy. To find the total kinetic energy related to the rotation of the body, the sum of the kinetic energy of every particle due to the rotational motion is taken.
Let us assume there are n particles, K.E. is represented as
Let us consider a body like a sphere, disc or a wheel rolling on a plane surface without slipping. A rolling body undergoes translational motion about the centre of mass and rotational motion about an axis passing through the centre of mass.
Total K.E. of a rolling body = Transitional K.E. + Rotational K.E.
Where m = mass of the body
R = Radius of the body
v = Translational speed of the body
I = M.I. of the body about an axis of rotation passing through the centre of mass.
K = Radius of gyration of the body
Radius of gyration of a given body about a given axis of rotation is the normal distance of a point from the axis, where if whole mass of the body is placed then its moment of inertia will be exactly same as it has with its actual distribution of mass. or
S.I. Unit of radius of gyration is metre.
The value of radius of gyration shall depend upon shape and size of the body, position and configuration of the axis of rotation and also on distribution of mass of the body w.r.t. the axis of rotation.
By comparing linear motion and rotational motion formulae we find that role played by moment of inertia in rotational motion is same as the role played by mass in translational motion.
In translational motion, mass of a body represents its inertia, ie reluctance to undergo a change in its state of rest or of uniform translational motion. Obviously, moment of inertial also represents inertia of a body in rotational motion (i.e. its reluctance to undergo a change in its state of rotation).Moment of inertia is also called rotational inertia.
Work done by a force F, acting on a particle of a body rotating about a fixed axis: the particle describes a circular path with centre C on the axis : arc gives the displacement of the particle.
Let’s assume mass m is moving with constant velocity along a line parallel to the x-axis, away form the origin. It’s angular momentum wrt origin remains constant. L =mvy
Both theorems are related to moment of inertia.
Parallel Axis Theorems | Perpendicular Axis Theorem |
---|---|
Moment of inertia of a body about a given axis I is equal to the sum of moment of inertia of the body about a parallel axis passing through its centre of mass I_{CM} and the product of mass of body (M) and square of normal distance d between two axes. | The sum of moment of inertia of a plane laminar body about two mutually perpendicular axes lying in its plane is equal to the moment of inertia about an axis passing through the point of intersection of these two axes and perpendicular to the plane of laminar body. |
If I_{x} and I_{y} be moment of inertia of the body about two perpendicular axes in its own plane and I_{z} be the moment of inertia about an axis passing through point O and perpendicular to the plane of lamina, then I_{z} = I_{x} + I_{y} | |
Applicable for any type of rigid body whether it is a two-dimensional or three-dimensional. | Applicable for laminar type / planer or two-dimensional bodies only. |
The point of intersection of the three axes (x,y and z) may be any point on the plane of body (it may even lie outside the body). This point may or may not be the center of mass of the body. |
Let us assume a small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of with respect to the initial position.
Now we need to find, which is this object?
In this scenario, Loss of Kinetic Energy = Gain in Potential Energy
; put given value of h in this equation.
This represents moment of inertia for Disc (with axis of rotation passing through C.G and perpendicular to its plane).
Body | Axis | Figure | I |
---|---|---|---|
Thin circular ring, radius R | Perpendicular to plane at centre | MR^{2} | |
Thin circular ring, radius R | Diameter | ||
Thin rod, length L | Perpendicular to rod, at mid point | ||
Circular disc, radius R | Perpendicular to disc at centre | ||
Circular disc, radius R | Diameter | ||
Hollow cylinder, radius R | Axis of cylinder | MR^{2} | |
Solid cylinder, radius R | Axis of cylinder | ||
Solid sphere, radius R | Diameter |
The moment of linear momentum of a given body about an axis of rotation is called as its angular momentum. If p be the linear momentum of a particle and r is its position vector from the point of rotation then .
When a body is projected at an angle with the horizontal in the uniform gravitational field of the earth, the angular momentum of the body about the point of projection increases, as it proceeds along the path.
Conservation of angular momentum and its applications:
According to the law of conservation of angular momentum, if no external unbalanced torque is acting on system , then total vector sum of angular momentum of different particles of the system remains constant.
If no external torque acts on a body, the angular momentum of the body remains constant.