Introduction
We come across many objects that follow rotational movements. A ceiling fan or a potter’s wheel, these rotating objects are a system of particles that consider the rotational motion. It is the combination of circular motion of large number of particles of a rigid system.
We know that force, energy and power are associated with rotational motion. These and other aspects of rotational motion are covered in this chapter. We shall see that all important aspects of rotational motion either have already been defined for linear motion.
Centre of Mass
Centre of mass is a very special point. The concept of centre of mass of a system enables us, in describing the overall motion of the system by replacing the system by an equivalent single point, where the entire mass of the body or system is supposed to be concentrated.
Now, suppose we have a system of n particles of masses m1, m2, m3 ...mn respectively along a straight line at distances x1, x2, x3 ...xn from the origin respectively. Then the centre of mass of the system is given by
Xcm=`\frac{m_1x_1+m_2x_2+...m_nx_n}{m_1+m_2+...m_n}`
Xcm=`\frac{m_1x_1+m_2x_2+...m_nx_n}{M}`
where M is the total mass of the system.
Centre of Gravity
The centre of gravity is that point of the body, where the whole weight of the body is supposed to be concentrated. We may define the centre of gravity of a body as that point where the total gravitational torque acting on the body is zero.
Now, suppose we have a system of n particles of masses w1, w2, w3 ...wn respectively along a straight line at distances x1, x2, x3 ...xn from the origin respectively. Then the centre of gravity of the system is given by
`X_{cd}=\frac{w_1x_1+w_2x_2+...w_nx_n}{w_1+w_2+...w_n}`
Recommended Books
- NCERT Textbook For Class 11 Physics Part 1 & 2
- CBSE All In One Physics Class 11 2022-23 Edition
- Oswaal CBSE Chapterwise Question Bank Class 11 Physics Book
- Modern's abc Plus of Physics for Class-11 (Part I & II)
Read also: Gravitation Class 11 Physics Notes Chapter 8
Motion of centre of mass
We can write equation of Centre of Mass for a system of particles as follow.
MRcm=`m_1r_1+m_2r_2+m_3r_3+...+m_nr_n`
Differentiating the two sides of the equation with respect to time, we get
`M\frac{dR}{dt}=m_1\frac{dr_1}{dt}+m_2\frac{dr_2}{dt}+...+m_n\frac{dr_n}{dt}`
The rate of change of position is velocity. So we can replace dR/dt with vcm where vcm is the velocity of the centre of mass.
Mvcm=`m_1v_1+m_2v_2+...+m_nv_n`
`M\frac{dv_{cm}}{dt}=m_1\frac{dv_1}{dt}+m_2\frac{dv_2}{dt}+...+m_n\frac{dv_n}{dt}`
Change in velocity is acceleration, so we get
Macm=`m_1a_1+m_2a_2+...+m_na_n`
where a1, a2, .....an are the acceleration of first, second, and ....nth particle respectively and acm is the acceleration of the centre of mass of the system of particles.
Linear Momentum of a System of Particles
The linear momentum of a particle is defined as
P=mv
and according to Newton's second law,
`F=\frac{dP}{dt}`
i.e. the rate of change of linear momentum of a particle is equal to the net force acting on the object. Using equation of motion of center of mass we can write
`Mv_{cm}=\sum_{i=1}^{n}m_iv_i`
`\sum_{i=1}^{n}P_{i}=\sum_{i=1}^{n}m_iv_i`
L.H.S. is the summation of linear momentum of n particles of the system, which is equal to product of the total mass of the system and velocity of the center of mass of the system.
`P=Mv_{cm}`
Differentiating the above equation w.r.t. time, we get
`\frac{dP}{dt}=M\frac{dv_{cm}}{dt}=ma_{cm}=F_{ext}`
Now, if the net external force on the system is zero, the linear momentum of the system, is conserved and the centre of mass will move with constant velocity.
Read also: Equilibrium Class 11 Notes Chemistry Chapter 7
Rigid Body
A rigid body is a collection of large number of particles, moving in a constrained manner. The constraint is that separation between any two particles of the system does not change.
Motion of Rigid Body
(i). Translation
If any line drawn on the rigid body remains parallel to itself throughout the motion, then the body is said to be in pure translation. In pure translation, all the particles of the body have equal velocity and acceleration at all instants and they cover equal distance and displacement in equal time.
(ii). Rotation
If any line drawn on the rigid body does not remain parallel to itself throughout its motion, then the body is said to be rotating. For example, the ceiling fan, bicycle wheel or a football rolling on ground.
(iii). Axis of Rotation
An imaginary line drawn perpendicular to the plane of motion of different points of the body and passing through the stationary point is called the axis of rotation.
Read also: Conceptual Questions for Class 11 Physics Chapter 7 System of Particles and Rotational Motion
(iv). Angle of Rotation (θ)
When the object rotates, its configuration changes, the angle by which any line drawn on the object rotates during the change in configuration is the angle of rotation.
While the body rotates, every point of the body moves in a circle, whose centre lies on axis of rotation, and every point experiences the same angular displacement during a particular time interval.
(v). Angular Velocity (ω)
The rate of rotation is measured by angular velocity. The angular velocity is defined as
`ω=\frac{dθ}{dt}`
The unit of the angular velocity is rad/s.
(v). Angular Acceleration (α)
The angular acceleration is defined as
`α=\frac{dω}{dt}`
The unit of the angular acceleration is rad/s2.
Dot Product and Cross Product
(i). The Dot Product of Two Vectors
The scalar product or dot product of any two vectors `\vec{A}` and `\vec{B}`, denoted as `\vec{A}` . `\vec{B}` (read as `\vec{A}` dot `\vec{B}`) is defined as
`\vec{A}` . `\vec{B}` = AB cosθ
Where A & B are magnitudes of vectors `\vec{A}` and `\vec{B}` respectively and θ is the smaller angle between them. Dot product is called scalar product as A, B and cosθ are scalars. Both vectors have a direction but their scalar product does not have a direction.
Properties
Dot product is commutative
A . B = B . ADot product is distributive
A . (B + C) = A . B + A . C
Dot product of a vector with itself gives square of its magnitude
A . A = AA cosθ = A
A . (λB) = λ(A . B)
where λ is a real number
- `\hat{i}` . `\hat{j}` = `\hat{j}` . `\hat{k}` = `\hat{k}` . `\hat{i}` = 0
- `\hat{i}` . `\hat{i}` = `\hat{j}` . `\hat{j}` = `\hat{k}` . `\hat{k}` = 1
(ii). The Cross Product of Two Vectors
The vector product or cross product of any two vectors `\vec{A}` and `\vec{B}`, denoted as `\vec{A}` x `\vec{B}` (read as `\vec{A}` cross `\vec{B}`) is defined as
`\vec{A}` x `\vec{B}` = AB sinθ
Where A & B are magnitudes of vectors `\vec{A}` and `\vec{B}` respectively and θ is the smaller angle between them. Cross product is called vector product as A, B and sinθ are scalars. Both vectors have a direction and their vector product has a same direction.
Properties
The vector product is do not have Commutative Property.
A×B = – (B×A)
The following property holds true in case of vector multiplication
(kA)×B= k(A×B) =A×(kB)
If the given vectors are collinear then
A×B= 0
Following the above property, We can say that the vector multiplication of a vector with itself would be
A×A= |A||A|sin0 `\hat{n}` = 0
Also in terms of unit vector notation
`\hat{i}×\hat{i}=\hat{j}×\hat{j}=\hat{k}×\hat{k}=0`
From the above discussion it also follows that
`\hat{i}×\hat{j}=\hat{k}=−\hat{j}×\hat{i}`
`\hat{j}×\hat{k}=\hat{i}=−\hat{k}×\hat{j}`
`\hat{k}×\hat{i}=\hat{j}=−\hat{i}×\hat{k}`
Relation between Angular Acceleration and Linear Acceleration
Angular Acceleration is given by
`α=\frac{dω}{dt}` .....(1)
Linear Acceleration is given by
`a=\frac{dv}{dt}` .....(2)
From eq (1) and (2),
`a=\frac{d(r\times ω)}{dt}`
`a=r\times\frac{dω}{dt}`
`a=r\times α`
Relation between Angular Velocity and Linear Velocity
Let, any rigid body is rotating about any rotational axis with angular velocity (ω). If distance of a particle at a perpendicular be r from the fixed axis and linear velocity be v of any particle of thye rigid system then relation between them is given by
`\vec{v}=\vec{r}\times \vec{ω}`
Torque
The tendency of a force to rotate the body to which it is applied is called torque. The torque, specified with regard to the axis of rotation, is equal to the magnitude of the component of the force vector lying in the plane perpendicular to the axis, multiplied by the shortest distance between the axis and the direction of the force component.
`\vec{τ}=\vec{r}\times\vec{F}`
Relation between Torque and Angular Velocity
`\vec{L}=\vec{r}\times\vec{P}`
Differentiating w.r.t. t, we get
`\frac{d\vec{L}}{dt}=\frac{d}{dt}(\vec{r}\times\vec{P})`
`\frac{d\vec{L}}{dt}=\vec{r}\times\frac{d\vec{P}}{dt}+\frac{d\vec{r}}{dt}\times\vec{P}`
`\frac{d\vec{L}}{dt}=\vec{r}\times \vec{F}+\vec{v}\times m\vec{v}`
`\frac{d\vec{L}}{dt}=\vec{r}\times \vec{F}+(\vec{v}\times \vec{v})`
`\frac{d\vec{L}}{dt}=\vec{r}\times \vec{F}+0` ... [`\therefore\vec{v}\times \vec{v}=0`]
`\frac{d\vec{L}}{dt}=\tau`
Moment of Inertia
When any object rotate about any axis then it has tendency to resist its motion, this tendency of resistance is called moment of inertia. It is denoted by I and it's SI unit is kg/m2.
Moment of Inertia of any object is defined as product of mass of that object and square of perpendicular distance of rotational axis.
`I=MR^2`
Theorems of Perpendicular and Parallel Axes
(i). Theorems of Perpendicular Axes
It states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.
Consider a lamina in x-y plane as shown in the figure and suppose that it consists of n particles of masses m1, m2, m3 ...mn at perpendicular distances r1, r2, r3 ...rn respectively from the axis OZ and suppose the corresponding perpendicular distances of these particles from the OY are x1, x2, x3 ...xn and from the axis OX are y1, y2, y3 ...yn respectively.
Let Ix, Iy, and Iz be the moment of inertia of the lamina about axes OX, OY and OZ respectively. Now,
`I_{x}=\sum_{i=1}^{n}m_iy_{i}^{2}` .....(1)
Similarly `I_{y}=\sum_{i=1}^{n}m_ix_{i}^{2}` .....(2)
and `I_{z}=\sum_{i=1}^{n}m_ir_{i}^{2}` .....(3)
Adding equations (i) and (ii), we get
`I_{x}+I_{y}=\sum_{i=1}^{n}m_{i}(y_{i}^{2}+x_{i}^{2})`
From the figure, we can see
`r_{i}^{2}=x_{i}^{2}+y_{i}^{2}`
`I_{x}+I_{y}=\sum_{i=1}^{n}m_{i}r_{i}^{2}=I_{z}`
`I_{z}=I_{x}+I_{y}`
(ii). Theorems of Parallel Axes
It states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
Consider a rigid body, as shown in the figure and suppose we know the moment of inertia of the body about axis AB, and want to find the moment of inertia of the body about EF which is at a perpendicular distance d from AB.
Suppose the rigid body is made up of n particles of masses m1, m2, m3 ... mn at perpendicular distances r1,r2, r3, ...rn respectively from the axis AB passing through the centre of mass C of the body. If ri is the perpendicular distance of the particle from the axes AB ; then
`I=\sum_{i=1}^{n}m_{i}r_{i}^{2}`
`I=\sum_{i=1}^{n}m_{i}(R+a)^{2}`
`I=\sum_{i=1}^{n}m_{i}(R^{2}+a^{2}+2Ra)`
`I=\sum_{i=1}^{n}m_{i}R^{2}+\sum_{i=1}^{n}m_{i}a^{2}+\sum_{i=1}^{n}m_{i}2Ra`
`I=\sum_{i=1}^{n}m_{i}R^{2}+\sum_{i=1}^{n}m_{i}a^{2}+0`
`I=I_{cm}+\sum_{i=1}^{n}m_{i}R^{2}`
Rolling motion
Rolling Motion of a body is a combination of both translational and rotational motion of a round shaped body placed on a surface. When a body is set in rolling motion, every particle of body has two velocities – one due to its rotational motion and the other due to its translational motion, and the resulting effect is the vector sum of both velocities at all particles.
Rolling Motion is classified in two categories – Pure Rolling and Rolling with Sliding. Pure rolling is a case when there is no relative motion at point of contact of rolling body and the surface; and body is considered to be rotating about this point of contact frame.
Rotational Kinetic Energy
If the mass of ith particle is mi and its speed is vi, its kinetic energy is
`K_{i}=\frac{1}{2}m_{i}v_{i}^{2}`
`K_{i}=\frac{1}{2}m_{i}r_{i}^{2} ω^2`
`K_{i}=\frac{1}{2}(m_{i}r_{i}^{2}) ω^2`
`K_{i}=\frac{1}{2}I ω^2`
Summary
- A rigid body is one for which the distances between different particles of the body do not change, even though there are forces on them.
- The centre of mass of a rigid body is defined as a point, where the entire mass of the rigid body is supposed to be concentrated and the nature of motion of the body shall remain unaffected, if all the forces acting on the body were applied directly on the centre of mass of the body.
- The centre of gravity of a body is that point where, the whole weight of the body is supposed to be concentrated.
- Angular displacement (θ) is the angle described by the position vector `\vec{r}` about the axis of rotation.
- Angular velocity (ω) is the rate at which angular displacement changes with time i.e. ω = dθ / dt.
- Angular acceleration (α) of a rigid body is the rate of change of angular velocity of the body about the given axis of rotation i.e. α = dω / dt.
- Angular momentum (`\vec{L}`) of a body about a given axis is the product of linear momentum and perpendicular distance of line of action of linear momentum vector from the axis of rotation i.e., L = r x P.
- Torque (τ): A pair of equal and opposite forces with different lines of action is known as a couple or torque. A couple produces rotation without translation. It can also be defined as the rate of change of angular momentum of a rigid body i.e. τ = dL / dt.
- Moment of Inertia (I) of a rigid body about an axis is defined as the sum of the product of masses of all the particles of the body and squares of respective perpendicular distances from the axis of rotation. I = Σm r2.
- Kinetic Energy of Rotation of a body is the energy possessed by the body on account of its rotation about a given axis.
- Radius of gyration (k) of a body about a given axis is the perpendicular distance of a point from the axis, where if whole mass of the body were concentrated.
- Theorem of perpendicular axis states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axis concurrent with perpendicular axis and lying in the plane of the body.
- Theorem of parallel axis states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
- In pure translation motion, every particle of the body moves with the same velocity at any instant of time.
- In rotational motion of a rigid body about a fixed axis, every particle of the rigid body moves in a circle, which lies in the plane perpendicular to the axis and has its centre on the axis.
- Every point on the rotating rigid body has same angular velocity.
- To determine the motion of the centre of mass of a system, we need to know only the external forces on the system.
- If the total forces acting on a system is zero, then the total linear momentum of the system is conserved.
- At the centre of gravity of a body, the total gravitational torque on the body is zero.
- The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary from one part of the body to the other.
- We can calculate the moment of inertia of a body about a given axis by using the perpendicular axes theorem and parallel axis theorem.
- For a rigid body, if the torque acting on the body is zero, the component of angular momentum about the fixed axis of such a rotating body is constant.
- Torque (τ) acting on a system of particles or rigid body vanishes if either F = 0 or r = 0 or the angle between them is 0° or 180°. And if the torque acting on the body is zero, then the angular momentum of the body is conserved.
