SYSTEM OF PARTICLES AND ROTATORY MOTION – PART 1

We have discussed the motion of a body by considering it as a point object. A point object means an object having a certain mass but negligible size. An extended object or a real object is made up of a large number of particular particles that exert force on each other.

System of particles

Collection of large number of particles interacting with each other is called system of particles.

Rigid Body

Rigid body is a body with a perfectly definite and unchanging shape. The distances between all pairs of particles of such a body do not change. In real situation no body is a perfectly rigid body. However, in bodies like wheels, steel beams, metallic sphere, wooden block etc. deformation under the force is so small that they can be considered as rigid bodies.

Motion of a rigid body

The motion of a rigid body which is not pivoted or fixed in some way is either a pure translation or a combination of translation and rotation. The motion of a rigid body which is pivoted or fixed in some way is rotation. The rotation may be about an axis that is fixed or moving.

Translational motion

At any instant of time if all the particles of a body have the same velocity, then the motion is said to be Translational.

Ex: Wooden block sliding on a inclined plane.

Rotational motion (Rotation)

If every particle of the body moves in a circle which lies in a plane perpendicular to the fixed axis and has its centre on the fixed axis, then the motion is said to be Rotational.

Ex: A ceiling fan

Note: Any particle lying on the axis of rotation remains at rest while the rigid body rotates about the axis of rotation. Thus, axis of rotation is fixed.

Precession

The movement of the axis of the rotating body around the vertical axis is termed as precession. While precession the point of contact of the rotating body with the ground is fixed.

Centre of mass

Centre of mass of a system of particles is the point where the entire mass of the system can be assumed to be concentrated.

Note: This point like mass has the same type of translational motion as the system as a whole if some net external force acts on this point like mass as acting on the system. Centre of mass of a body or a system is its balancing point.

Centre of mass of a two particle system

Consider two particles on the X-axis. Let the distances of the two particles be 𝑥₁ and 𝑥₂ respectively from origin. The masses of the two particles be m₁ and m₂ respectively. The centre of mass of the system which is at a distance X_cm from

𝑂 is given by,

𝑿_𝒄𝒎 = (𝒎_𝟏𝒙_𝟏 + 𝒎_𝟐𝒙_𝟐)/(𝒎_𝟏 + 𝒎_𝟐)

If 𝑅 be the position vector of the centre of mass then

𝑅 = 𝑋_𝑐𝑚 𝑖̂

Note:

(1) If the two particles have the same mass 𝑚₁ = 𝑚₂ = 𝑚, then 𝑋_𝑐𝑚 = (𝑚𝑥₁ + 𝑚𝑥₂)/(𝑚 + 𝑚) = 2m(𝑥₁ + 𝑥₂)/2𝑚

𝑋_𝑐𝑚 = (𝑥₁ + 𝑥₂)/2

Thus for two particles of equal mass the centre of mass lies exactly midway between them.

(2) If the particles not lying in the straight line, we define x and y axes in the plane in which the particle lie and represent the positions of the two particle by coordinates (𝑥₁, 𝑦₁) and (𝑥₂, 𝑦₂) respectively. The centre of mass located by the co-ordinates (𝑋_𝑐𝑚, 𝑌_𝑐𝑚) is given by, 𝑿_𝒄𝒎 = (𝒎_𝟏𝒙_𝟏 + 𝒎_𝟐𝒙_𝟐)/(𝒎_𝟏 + 𝒎_𝟐) and 𝒀_𝒄𝒎 = (𝒎_𝟏𝒚_𝟏 + 𝒎_𝟐𝒚_𝟐)/(𝒎_𝟏 + 𝒎_𝟐).

If 𝑅 be the position vector of the centre of mass then 𝑅 = 𝑋_𝑐𝑚 𝑖̂+ 𝑌_𝑐𝑚𝑗̂

Centre of mass of three particle system

Consider three particles, not lying in same straight line. Let the masses of these particle be 𝑚₁, 𝑚₂ and 𝑚₃ respectively.

Positions of the three particles are represented by co-ordinates (𝑥₁, 𝑦₁), (𝑥₂, 𝑦₂) and (𝑥₃, 𝑦₃). The centre of mass located by the co-ordinates (𝑋_𝑐𝑚 , 𝑌_𝑐𝑚) is given by,

𝑋_𝑐𝑚 = (𝑚₁𝑥₁ + 𝑚₂𝑥₂ + 𝑚₃𝑦₃)/(𝑚₁ + 𝑚₂ + 𝑚₃) and 𝑌_𝑐𝑚 = (𝑚₁𝑦₁ + 𝑚₂𝑦₂ + 𝑚₃𝑦₃)/(𝑚₁ + 𝑚₂ + 𝑚₃)

𝑅 = 𝑋_𝑐𝑚 𝑖̂+ 𝑌_𝑐𝑚𝑗̂

Note: For the particles of equal masses, 𝑚₁ = 𝑚₂ = 𝑚₃ = 𝑚

𝑋_𝑐𝑚 = (𝑚𝑥₁ + 𝑚𝑥₂ + 𝑚𝑥₃)/(𝑚 + 𝑚 + 𝑚) = 𝑚(𝑥₁ + 𝑥₂ + 𝑥₃)/3𝑚

𝑋_𝑐𝑚 = (𝑥₁ + 𝑥₂ + 𝑥₃)/3

𝑌_𝑐𝑚 = (𝑚𝑦₁ + 𝑚𝑦₂ + 𝑚𝑦₃)/(𝑚 + 𝑚 + 𝑚) = 𝑚(𝑦₁ + 𝑦₂ + 𝑦₃)/3𝑚

𝑌_𝑐𝑚 = (𝑦₁ + 𝑦₂ + 𝑦₃)/3

Thus for three particles of equal mass, the centre of mass coincides with the centroid of the triangle formed by the particles.

Centre of mass of a system of n-particles

Consider a system of n-particles. Let 𝑚₁, 𝑚₂ … … … 𝑚_𝑛 be the respective masses of the particles. Let 𝑋_𝑐𝑚, 𝑌_𝑐𝑚 and 𝑍_𝑐𝑚 are the co-ordinates of the centre of mass of the system.

Then 𝑋_𝑐𝑚 = (𝑚₁𝑥₁ + 𝑚₂𝑥₂+. … … … … + 𝑚_𝑛𝑥_𝑛)/(𝑚₁ + 𝑚₂ + … … … … . . +𝑚_𝑛)

Similarly

and

The position vector of the centre of mass is, 𝑹 = 𝑿_𝒄𝒎𝒊+ 𝒀_𝒄𝒎j + 𝒁_𝒄𝒎k

Then

Note: Centre of mass of a system may or may not lie inside the system.

Centre of mass of a rigid body

For system of n − particles the centre of mass is given by,

This equation is applicable to rigid body also. In case of a Rigid body the number of particles is so large that it is impossible to carry out the summation over individual particles in the equation. Since the spacing of the particle is small, we can treat the body as a continuous distribution of mass. We subdivide the body into n small elements of mass dm₁, dm₂, ………dm_n. Hence sum

can be replaced by the integral ∫ 𝑑𝑚 𝑟 .

Where 𝑅 = 𝑋_𝑐𝑚 𝑖+ 𝑌_𝑐𝑚j + 𝑍_𝑐𝑚k

Now 𝑋_𝑐𝑚 𝑖+ 𝑌_𝑐𝑚j + 𝑍_𝑐𝑚k = (1/𝑀) ∫(𝑥i+ yj + 𝑧k) 𝑑𝑚

Comparing the co-efficient of I, j and k

𝑋_𝑐𝑚 = (1/𝑀)∫𝑥𝑑𝑚

𝑌_𝑐𝑚 = (1/𝑀)∫y𝑑𝑚

𝑍_𝑐𝑚 = (1/𝑀)∫z𝑑𝑚

If we choose, the centre of mass as the origin of our co-ordinate system, then, 𝑅 = 0 implies that, ∫𝑟𝑑𝑚 = 0 and ∫𝑥𝑑𝑚 = ∫𝑦𝑑𝑚 = ∫𝑧𝑑𝑚 = 0

Important conclusions:

i. In case of homogeneous bodies like a circular solid disc, an ice cube or a sugar cube, solid sphere, hollow sphere, a marble ball, a billiard ball, an iron ball uniform thin rod etc. The centre of mass coincides with the geometric centers of the bodies.

 ii. In the case of bodies having axis of symmetry like a solid cylinder, hollow cylinder a wheel etc, the centre of mass lies on the axis of symmetry of the body.

Centre of mass of uniform rod

Consider a uniform rod of length L. Let one end A of the rod is taken as the origin. Since the rod is uniform, the mass per unit length of the rod (𝜆) is constant.

Consider a small element of the rod of length dx at a distance 𝑥 from end 𝐴 and having the mass, 𝑑𝑚 = 𝜆𝑑𝑥. The co − ordinate of the centre of the rod is given by,

Motion of centre of mass

Consider a system having n-particles of masses 𝑚₁, 𝑚₂, 𝑚₃ … … . 𝑚𝑛. Let 𝑟₁ , 𝑟₂ , 𝑟₃ … 𝑟_𝑛 be their respective position vectors. The centre of mass of the system is given by,

𝑀𝑅 = 𝑚₁𝑟₁ + 𝑚₂𝑟₂ + 𝑚₃𝑟₃+. … … . +𝑚_𝑛𝑟_𝑛

Differentiating the two sides of the equation with respect to time,

𝑀(𝑑𝑅/𝑑𝑡) = 𝑚₁(d𝑟₁/dt) + 𝑚₂(d𝑟₂/dt) + 𝑚₃(d𝑟₃/dt) +. … … . + 𝑚_n(d𝑟_n/dt)

𝑀𝑉 = 𝑚₁v₁ + 𝑚₂v₂ + 𝑚₃v₃+. … … . +𝑚_𝑛v_𝑛

Differentiating the two sides of the equation with respect to time again

𝑀(𝑑V/𝑑𝑡) = 𝑚₁(dv₁/dt) + 𝑚₂(dv₂/dt) + 𝑚₃(dv₃/dt) +. … … . + 𝑚_n(dv_n/dt)

𝑀A = 𝑚₁a₁ + 𝑚₂a₂ + 𝑚₃a₃+. … … . +𝑚_𝑛a_𝑛

Using Newton’s second law, 𝐹 = 𝑚𝑎

𝑀𝐴 = 𝐹₁ + 𝐹₂ + 𝐹₃ +. … … 𝐹_𝑛

𝑴𝑨 = 𝑭_𝒆𝒙𝒕 where 𝐹_𝑒𝑥𝑡 = 𝐹₁ + 𝐹₂ + 𝐹₃ +. … … 𝐹_𝑛

From the above equation we can conclude that, the centre of mass of the system of particles moves as if the mass of the system was concentrated at the centre of mass and all the external force were applied at that point.

Linear momentum of a system of particles

Consider a system of n particles of masses m₁, m₂, m₃, …. m_n moving with velocities 𝑣₁, 𝑣₂, 𝑣₃ …….. 𝑣_𝑛 respectively. The momentum of the system is given by,

𝑃 = 𝑝₁ + 𝑝₂ + 𝑝₃ +. … + 𝑝_n

𝑃 = 𝑚₁𝑣₁ + 𝑚₂𝑣₂ + 𝑚₃𝑣₃+. … … . + 𝑚_n𝑣_n

But we have,

𝑀𝑉 = 𝑚₁𝑣₁ + 𝑚₂𝑣₂ + 𝑚₃𝑣₃+. … … . + 𝑚_n𝑣_n

𝑀𝑉 = 𝑃 or 𝑃 = 𝑀𝑉

Thus the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of the centre of mass. Differentiating equation with respect to time

𝑑𝑃/𝑑𝑡 = 𝑀(𝑑𝑉/𝑑𝑡)

𝑑𝑃/𝑑𝑡 = 𝑀𝐴

But 𝑀𝐴 = 𝐹_𝑒𝑥𝑡

𝑑𝑃/𝑑𝑡 = 𝐹_𝑒𝑥𝑡

This is the Newton’s second law for system of particles. If 𝐹_𝑒𝑥𝑡 = 0 then, 𝑑𝑃/𝑑𝑡 = 0 implies that 𝑃 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Thus, when the total external force acting on a system of particles is zero, the total linear momentum of the system is constant. This is the law of conservation of the linear momentum of a system of particles.

𝐍𝐨𝐭𝐞: Now by considering equation, 𝑑𝑃/𝑑𝑡 = 0

𝑑(𝑀𝑉)/dt = 0

𝑀𝑑𝑉/𝑑𝑡 = 0

But 𝑀 ≠ 0 therefore 𝑑𝑣/𝑑𝑡 = 0, implies that 𝑉 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

This shows that, when the total external force on the system is zero, the velocity of the center of mass remains constant. In other words, a system interacting internally cannot accelerate itself.

Examples for motion of center of mass

(1) In Radio-active decay the process is caused by the internal force of the system. Therefore initial and final momentums are zero. The Nucleus _zX^A decays into a new nucleus _z-2Y^A-4 and a Helium Nucleus as shown in figure. Since the disintegration occurs only due to the internal force, the nucleus _z-2Y^A-4 and Helium nucleus must move in such a directions that sum of their momentum is zero and their center of mass moves along the path followed by the nucleus _zX^A before decay. However, when the decay of the nucleus is observed from a frame of reference with respect to which the nucleus is at rest, then the decay products fly off in the opposite directions. The Centre of mass of the system remains at rest. The heavy mass moves with less speed than that of the light mass.

(2) Explosion of a projectile in mid air (fire cracker).

Let us consider a projectile which explodes in air. Before explosion, the projectile moves along a parabolic path. After explosion each fragment moves along their own parabolic path but the centre of mass of the projectile continues to move in the same parabolic path.

Vector product or Cross product of two vectors

The vector product of two vectors gives a vector quantity. Definition: The vector product of two vectors is a single vector whose magnitude is equal to the product of the magnitude of two given vectors multiplied by the sine of the smaller angle between the two vectors and the direction of the vector is perpendicular to the plane containing the two vectors.

Explanation: Consider two vectors 𝐴 and 𝐵 such that the angle between them is 𝜃 then, the cross product of the vectors 𝐴 and 𝐵 is 𝐴 × 𝐵 (𝐴 𝑐𝑟𝑜𝑠𝑠 𝐵) which is given by, 𝑨⃗ × 𝑩⃗ = (𝑨𝑩 𝐬𝐢𝐧 𝜽)𝒏̂ 𝑛̂ is the unit vector which gives the direction of the vector 𝐴 × 𝐵.

𝒏̂ = 𝑨⃗ × 𝑩⃗ /𝑨𝑩 𝐬𝐢𝐧 𝜽 = 𝑨⃗ × 𝑩⃗/|𝑨⃗ × 𝑩⃗⃗ |

The direction of the vector 𝐴 × 𝐵 can be determined by right hand screw rule.

Right hand screw rule

Take a right handed screw with its head lying in the plane of 𝐴 and 𝐵 and the screw perpendicular to this plane. If we turn the head of the screw in the direction from A to B through a small angle 𝜃, then the tip of the screw advances in the direction of the vector 𝐴 × 𝐵,

Difference between Scalar product and vector product

Scalar product	Vector product
Scalar product of two vector is a scalar	Vector product of two vector is a vector
Scalar product is commutative	Vector product is not commutative
Scalar product of two equal or parallel vector is not equal to zero	Vector product of two equal or parallel vector is equal to zero
Scalar product of two perpendicular vectors is equal to zero	Vector product of two perpendicular vectors is not equal to zero

Properties of cross product

(i) The vector product is not commutative.

𝐴 × 𝐵 = (𝐴𝐵 sin 𝜃)𝑛̂

𝐵 × 𝐴 = (𝐵𝐴 sin 𝜃)(−𝑛̂) = −(𝐴𝐵 sin 𝜃)𝑛̂

𝐴 × 𝐵 = −(𝐵 × 𝐴 )

𝐴 × 𝐵⃗ ≠ 𝐵⃗ × 𝐴

Note: Angle between 𝐴 × 𝐵 and 𝐵 × 𝐴 is 180° or 𝜋 radian.

(ii) Vector product is distributive over vector addition.

𝐴 × (𝐵 + 𝐶 ) = [𝐴(𝐵 + 𝐶) sin 𝜃]𝑛̂

𝐴 × (𝐵 + 𝐶 ) = [𝐴𝐵 sin 𝜃 + 𝐴𝐶 sin 𝜃]𝑛̂

𝐴 × (𝐵 + 𝐶 ) = 𝐴𝐵 sin 𝜃𝑛̂ + 𝐴𝐶 sin 𝜃 𝑛̂

𝐴 × (𝐵 + 𝐶 ) = 𝐴 × 𝐵 + 𝐴 × 𝐶

Vector product of two parallel vectors or equal vectors

The angle between the parallel or equal vectors in zero.

𝐴 × 𝐵 = (𝐴𝐵 sin 𝜃)𝑛̂

𝐴 × 𝐵 = (𝐴𝐵 sin )𝑛̂ (sin 0 = 0)

𝐴 × 𝐵 = 0

For equal vectors, 𝐴 × 𝐴 = (𝐴𝐴 sin 𝜃)𝑛̂

𝐴 × 𝐴 = (𝐴𝐴 sin 0)𝑛̂

𝐴 × 𝐴 = 0

Similarly 𝑖̂× 𝑖̂ = (1 × 1 sin 0)𝑛̂ = 0

𝑗̂× 𝑗̂= (1 × 1 sin 0)𝑛̂ = 0

𝑘̂ × 𝑘̂ = (1 × 1 sin 0)𝑛̂ = 0

𝑖̂× 𝑖̂= 𝑗̂× 𝑗̂= 𝑘̂ × 𝑘̂ = 0

Vector product of two perpendicular vectors

The angle between two perpendicular vectors is 90°

𝐴 × 𝐵 = (𝐴𝐵 sin 𝜃)𝑛̂

𝐴 × 𝐵 = (𝐴𝐵 sin 90°)𝑛̂ (sin 90° = 0)

𝐴 × 𝐵 = (𝐴𝐵)𝑛̂

Similarly 𝑖̂× 𝑗̂= (1 × 1 sin 90°)𝑘̂ = 𝑘̂

𝑗̂× 𝑘̂ = (1 × 1 sin 90°)𝑖̂= 𝑖̂

𝑘̂ × 𝑖̂= (1 × 1 sin 90°)𝑗̂= 𝑗̂

𝑖̂× 𝑗̂= 𝑘̂, 𝑗̂× 𝑘̂ = 𝑖,̂ 𝑘̂ × 𝑖̂= 𝑗̂

Now 𝑗̂× 𝑖̂= (1 × 1 sin 90°)(−𝑘̂) = −𝑘̂

𝑘̂ × 𝑗̂= (1 × 1 sin 90°)(−𝑖̂) = −𝑖̂

𝑖̂× 𝑘̂ = (1 × 1 sin 90°) (−𝑗̂) = −𝑗̂

𝑗̂× 𝑖̂= −𝑘̂, 𝑘̂ × 𝑗̂= −𝑖,̂ 𝑖̂× 𝑘̂ = −𝑗̂

Cross – product of any two unit vectors in anticlockwise direction gives the positive value of the third unit vector. If the cross product of two unit vectors is taken in clock wise direction then it gives negative value of third unit vector.

Vector product of two vectors in their rectangular components (Analytical method)

The vector 𝐴 and 𝐵 can be written in terms of their rectangular component as, 𝐴 = 𝐴_𝑥𝑖̂+ 𝐴_𝑦𝑗̂+ 𝐴_𝑧𝑘̂

𝐵 = 𝐵_𝑥𝑖̂+ 𝐵_𝑦𝑗̂+ 𝐵_𝑧𝑘̂

𝐴 × 𝐵 = (𝐴_𝑥𝑖̂+ 𝐴_𝑦𝑗̂+ 𝐴_𝑧𝑘̂) × (𝐵_𝑥𝑖̂+ 𝐵_𝑦𝑗̂+ 𝐵_𝑧𝑘̂)

𝐴 × 𝐵 = 𝐴_𝑥𝐵_𝑥 (𝑖̂× 𝑖̂) + 𝐴_𝑥𝐵_𝑦 (𝑖̂× 𝑗̂) + 𝐴_𝑥𝐵_𝑧 (𝑖̂× 𝑘̂) + 𝐴_𝑦𝐵_𝑥 (𝑗̂× 𝑖)̂ + 𝐴_𝑦𝐵_𝑦 (𝑗̂× 𝑗̂) + 𝐴_𝑦𝐵_𝑧(𝑗̂× 𝑘̂) +𝐴_𝑧𝐵_𝑥(𝑘̂ × 𝑖)̂ + 𝐴_𝑧𝐵_𝑦(𝑘̂ × 𝑗̂) + 𝐴_𝑧𝐵_𝑧(𝑘̂ × 𝑘̂)

But 𝑖̂× 𝑖̂= 𝑗̂× 𝑗̂= 𝑘̂ × 𝑘̂ = 0 𝑖̂× 𝑗̂= 𝑘̂, 𝑗̂× 𝑘̂ = 𝑖,̂ 𝑘̂ × 𝑖̂= 𝑗̂ and 𝑗̂× 𝑖̂= −𝑘̂, 𝑘̂ × 𝑗̂= −𝑖,̂ 𝑖̂× 𝑘̂ = −𝑗̂ 𝐴 × 𝐵 = (𝐴_𝑥𝐵_𝑦)𝑘̂ + (𝐴_𝑥𝐵_𝑦)(−𝑗̂) + (𝐴_𝑦𝐵_𝑥)(−𝑘̂) + (𝐴_𝑦𝐵_𝑧)(𝑖)̂ + (𝐴_𝑧𝐵_𝑥)(𝑗̂) + (𝐴_𝑧𝐵_𝑦)(−𝑖)̂

𝑨⃗ × 𝑩⃗ = (𝑨_𝒚𝑩_𝒛 − 𝑨_𝒛𝑩_𝒚)𝒊̂+ (𝑨_𝒛𝑩_𝒙 − 𝑨_𝒙𝑩_𝒛)𝒋̂+ (𝑨_𝒙𝑩_𝒚 − 𝑨_𝒚𝑩_𝒙)𝒌̂

Vector product of two vectors in their rectangular components (Determinant method)

𝐴 × 𝐵 = 𝑖̂(𝐴_𝑦𝐵_𝑧 − 𝐴_𝑧𝐵_𝑦) − 𝑗̂(𝐴_𝑥𝐵_𝑧 − 𝐴_𝑧𝐵_𝑥) + 𝑘̂(𝐴_𝑥𝐵_𝑦 − 𝐴_𝑦𝐵_𝑥)

𝐴 × 𝐵 = (𝐴_𝑦𝐵_𝑧 − 𝐴_𝑧𝐵_𝑦)𝑖̂+ (𝐴_𝑧𝐵_𝑥 − 𝐴_𝑥𝐵_𝑧)𝑗̂+ (𝐴_𝑥𝐵_𝑦 − 𝐴_𝑦𝐵_𝑥)𝑘̂

Magnitude of A⃗ × B⃗ is given by, |𝑨⃗ × 𝑩⃗| = √(𝑨_𝒚𝑩_𝒛 − 𝑨_𝒛𝑩_𝒚)^𝟐 + (𝑨_𝒛𝑩_𝒙 − 𝑨_𝒙𝑩_𝒛)^𝟐 + (𝑨_𝒙𝑩_𝒚 − 𝑨_𝒚𝑩_𝒙)^𝟐

Angular displacement

It is defined as angle described by the radius vector in given time.

𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡(∆𝜃) = 𝐴𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ/𝑅𝑎𝑑𝑖𝑢𝑠 = 𝑥/𝑟

SI unit of angular displacement is radian (rad). Angular displacement is dimensionless quantity.

Angular velocity

It is defined as the ratio of the angular displacement of the particle to the time interval for this displacement.

Explanation: Consider a particle of the body whose position at any instant is P. Let after time ∆t the position of the particle on the circle in which it is moving is P’. If ∆𝜃 is the angular displacment of the particle in time ∆t then average angular velocity of the rotating particle is given by, 𝜔 = ∆𝜃/∆𝑡. As ∆t tends to zero the ratio (∆𝜃/∆𝑡) approaches a limit which is instantaneous angular velocity of the particle.

𝝎 = 𝐥𝐢𝐦 _{∆𝒕→𝟎} ∆𝜽/∆𝒕 = 𝒅𝜽/𝒅𝒕

The direction of angular velocity is along the axis of rotation which can be determined by Right[1]hand screw rule. S.I unit of 𝜔 is radian/second (rads^-1) and Dimensional formula is [M⁰L⁰T^-1].

Note: If a body is rotating in the direction of Increasing 𝜃 (anticlockwise) then angular velocity of the body is positive. If the body is rotating in a direction of decreasing 𝜃 (clockwise) then angular velocity of the body is negative.

Vector relation between linear velocity and angular velocity

Consider a particle at position P of a rigid body. As the body rotates the particle also moves from position P to position P’ Let its linear displacement 𝑃𝑃′ = 𝑑𝑟 and its angular displacement is 𝑑𝜃. Now 𝑑𝑟 = 𝑑𝜃 × 𝑟_⊥. Dividing both sides by dt, 𝑑𝑟/𝑑𝑡 = 𝑑𝜃/𝑑𝑡 × 𝑟_⊥

𝑣 = 𝜔⃗ × 𝑟_⊥

But from the ∆OPC, 𝑟_⊥ = 𝑟 − 𝑂𝐶

Substituting for 𝑟_⊥, 𝑣 = 𝜔 × (𝑟 − 𝑂𝐶)

𝑣 = 𝜔 × 𝑟 − 𝜔 × 𝑂𝐶

But 𝜔 × 𝑂𝐶 = 0 as 𝜔 is along 𝑂𝐶 ∴ 𝑣 = 𝜔 × 𝑟 − 0

𝒗 = 𝝎 × 𝒓 where 𝑟 → position vector of the particle at P. This relation shows that the linear velocity of particles situated at different position from axis of rotation is different.

Note: (i) For the particle laying on the axis of rotation 𝑟 = 0. Therefore the linear velocity of the particle at the axis of rotation is zero. (ii) Angular velocity of every particle of the rigid body is same as that of the angular velocity of the rigid body this is because every particle in the rigid body rotates through the same angle in the same interval of time.

Angular Acceleration

The angular acceleration can be defined as the time rate of change of angular velocity.

Explanation: Let 𝜔₁ and 𝜔₂ be the angular velocities of the body at instants t₁ and t₂ respectively. Then the average angular acceleration of the body is given by, 𝛼̅ = (𝜔₂ − 𝜔₁)/(𝑡₂ − 𝑡₁) = ∆𝜔/∆𝑡.

The instantaneous angular acceleration is given by, 𝛼 = lim _∆𝑡→0 ∆𝜔/∆𝑡 = 𝑑𝜔/𝑑𝑡 𝜶 = 𝒅𝝎/𝒅𝒕 = 𝒅/𝒅𝒕(𝒅𝜽/𝒅𝒕) = 𝒅^𝟐𝜽/𝒅𝒕^𝟐

The unit of angular acceleration is rads^-2.Dimensional formula is [𝑀⁰𝐿⁰𝑇⁻²] Note: If the axis of rotation is fixed the direction of 𝜔 and hence that of 𝛼 is fixed, then vector equation reduces to scalar equation, 𝜶 = 𝒅𝝎/𝒅t

Moment of force or Torque

Torque acting on a particle is defined as the product of the magnitude of the force acting on the particle and the perpendicular distance of the application of force from the axis of rotation of the particle.

Explanation: Consider a particle at P in X-Y plane with position vector 𝑟. Let 𝐹 acts on the particle at angle 𝜃 with the direction of the position vector, then torque, 𝜏 acting on the particle with respect the origin is given by, 𝝉⃗ = 𝒓⃗ × 𝑭⃗ The magnitude of torque 𝜏 is given by, 𝝉 = 𝒓𝑭 𝐬𝐢𝐧 𝜽 = 𝑭( 𝒓 𝐬𝐢𝐧 𝜽) = 𝑭𝒓_⊥ where 𝑟_⊥ = 𝑟 sin 𝜃 is perpendicular distance of the line of 𝐹 from O. Unit of torque is Nm (newton-metre). Dimensional formula is [ML²T^-2].

Direction of torque

The direction of torque is perpendicular to the plane containing 𝑟 and 𝐹 and can be determined by Right hand rule.

Explanation: When a Force 𝐹 is applied at a distance r from the bolt in anticlockwise, the bolt moves up, thus the torque acting on the bolt is in the upward direction. On the other hand when force 𝐹 is applied at a distance r from the bolt in clockwise direction the bolt moves downward, thus the torque acting on the bolt is in the down ward direction.

Maximum and minimum values of torque

1. If 𝜃 = 0 the force acts in the direction of position vector, then 𝜏 = 𝑟𝐹 sin 0 = 0 2. If 𝜃 = 90° Force acts perpendicular to the position vector then 𝜏 = 𝑟𝐹 sin 90° = 𝑟𝐹

Note: 𝑟 × 𝐹 is a vector product.

Therefore properties of a vector product of two vectors apply to it. If the direction of 𝐹 is reversed, the direction of torque is reversed. If direction of both 𝑟 and 𝐹 are reversed. The direction of the torque remains same.

Angular momentum (moment of momentum)

Angular momentum of a particle about an axis of rotation is defined as the product of linear momentum of the particle and the perpendicular distance of the particle from the axis of rotation.

Explanation: Consider a particle at P of mass m moving with a velocity 𝑣 in a circular path about z-axis. The angular momentum is, 𝒍 = 𝒓⃗ × 𝒑⃗ The magnitude of the angular momentum vector is, 𝒍 = 𝒓 𝒑 𝐬𝐢𝐧 𝜽 where 𝜃 is the angle between 𝑟 and p.

The direction of angular momentum is perpendicular to the plane containing 𝑟 and 𝑃. Unit of angular momentum is kgm²s^-1 and Dimensional formula is [ML²T^-1]

Note: 𝑙 = 𝑟(𝑝 sin 𝜃) = 𝑝(𝑟 sin 𝜃) where 𝑝 sin 𝜃 → The component of momentum in the direction perpendicular to 𝑟

𝑟 sin 𝜃 → Perpendicular distance of the directional line of 𝑃 from the origin

Relation between Torque and Angular momentum of a particle

We know, 𝑙 = 𝑟 × 𝑝

Differentiating both sides with respect to 𝑡 we have 𝑑𝑙/𝑑𝑡 = 𝑑/𝑑𝑡(𝑟 × 𝑝)

𝑑𝑙/𝑑𝑡 = 𝑟 × (𝑑𝑝/𝑑𝑡) + (𝑑𝑟/𝑑𝑡) × 𝑝

𝑑𝑙/𝑑𝑡 = 𝑟 × 𝐹 + 𝑣 × (𝑚𝑣)

𝑑𝑙/𝑑𝑡 = 𝑟 × 𝐹 + 𝑚(𝑣 × 𝑣)

𝑑𝑙/𝑑𝑡 = 𝑟 × 𝐹 ∵ (𝑣 × 𝑣 = 0)

𝒅𝒍/𝒅𝒕 = 𝝉

Thus, the time rate of change of angular momentum of a particle is equal to the torque acting on it.

Note: This is the rotational analogue of the equation 𝐹 = 𝑑𝑝/𝑑𝑡 which expresses Newton’s second law for translational motion of a particle.

Torque and angular momentum for a system of particles

Consider a system of n-particles. Let 𝑙₁, 𝑙₂, 𝑙₃ … . . 𝑙_𝑛 be the angular moments of the particles of the system respectively about the origin O. The angular momentum of the system of particles is given by, 𝐿 = 𝑙₁ + 𝑙₂ + 𝑙₃ + ⋯ + 𝑙_𝑛

Differentiating with respect to 𝑡,

We have separated the contribution of the external and internal torques to the total (net) torque.

The contribution of internal force to the total torque on the system is zero, because the forces between any two particles of the system are equal and opposite, and these forces are directed along the line joining the two particles.

Thus, the time rate of change of angular momentum of a system of particles is equal to the net external torque acting on the system.

Conservation of angular momentum

If the total external torque on the system of particles is zero, the total angular momentum of the system of particles does not change with time.

𝐄𝐱𝐩𝐥𝐚𝐧𝐚𝐭𝐢𝐨𝐧:

We have dL/dt = 𝜏_𝑒𝑥𝑡

If 𝜏_𝑒𝑥𝑡 = 0, then 𝑑𝐿/𝑑𝑡 = 0 and 𝐿 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

This is the law of conservation of angular momentum.

Equilibrium of a rigid body

A rigid body is said to be in mechanical equilibrium, if both its linear momentum and angular momentum are not changing with time.

Conditions for equilibrium

(i) We have

If

then 𝑑𝑝/𝑑𝑡 = 0 and 𝑝 is constant If the total force on the body is zero then the total linear momentum of the body does not change with time. This is the condition for translational equilibrium.

(ii) We have

If

then 𝑑𝐿/𝑑𝑡 = 0 and 𝐿 is constant. If the total torque on the rigid body is zero, the total angular momentum of the body does not change with time. This is the condition for rotational equilibrium of the body.

Partial equilibrium

A body may be in translational equilibrium and not in rotational equilibrium or it may be in rotational equilibrium and not in translational equilibrium. Then the body is said to be in partial equilibrium. Ex: (i) consider a rod AB of negligible mass and length L. Let two equal and parallel force acts on the two ends of the rod as shown.

Since the forces are parallel, net force on the rod is ∑𝐹 = 𝐹 = 2𝐹 ≠ 0.

Hence the rod is not in the translational equilibrium. But the torque at 𝐴 is, 𝜏₁ = 𝐹 × 𝐿₂ which tends to rotate the rod anticlockwise. The torque at 𝐵 is, 𝜏₂ = 𝐹 × 𝐿₂ which tends to rotates the rod clockwise. Hence the net torque on the rod is zero. So the rod is in rotational equilibrium.

(ii) Now the force at B in the figure is reversed. Now we have same rod with two equal and opposite force applied perpendicular to the rod. Now the torque at 𝐴 is, 𝜏₁ = 𝐹 × 𝐿₂ in anticlockwise direction. at 𝐵, 𝜏₂ = 𝐹 × 𝐿₂ in clockwise direction. Net torque is 𝜏₁ + 𝜏₂ ≠ 0 But Force at A is exactly equal force at B and in opposite direction. So net force is = 𝐹 + (– 𝐹) = 0. Here the rod is in translational equilibrium but not in rotational equilibrium.

Couple

Two equal and opposite forces with different lines of action is known as couple. When a couple acts on a body, the body is in translational equilibrium but not in rotational equilibrium. Thus a couple rotates the body.

Ex:

(i) A tap is opened or closed when our figures apply a couple on it.

(ii) A lid of a bottle is also opened and closed when our fingers apply a couple on it.

Principle of moments

Consider two forces F₁ and F₂ parallel each other and perpendicular to the rod. Let these forces act on the rod at distances d₁ and d₂ respectively from the fulcrum. Let 𝑅 be the reaction of the support at fulcrum which is directed opposite to the forces F₁ and F₂. For rotational equilibrium,

𝑑₁𝐹₁ + (−𝑑₂𝐹₂) = 0

𝑑₁𝐹₁ − 𝑑₂𝐹₂ = 0

𝒅_𝟏𝑭_𝟏 = 𝒅_𝟐𝑭_𝟐

Anticlockwise moment of force = clockwise moment of force. This is known as the principle of moments.

Note: For translation equilibrium, 𝑅 + (−𝐹₁) + (−𝐹₂) = 0

𝑅 − 𝐹₁ − 𝐹₂ = 0

𝑅 = 𝐹₁ + 𝐹₂

Lever

An ideal lever is a light rod pivoted at a point along its length. It works on the principle of moments.

Ex: A see- saw on the children’s playground.

Explanation: We have 𝑑₁𝐹₁ = 𝑑₂𝐹₂

In the case of lever Force F₁ is known as load and force F₂ is known as effort. Distance from the fulcrum d₁ is called load arm. Distance d₂ is called as effort arm Load

𝒂𝒓𝒎 × 𝒍𝒐𝒂𝒅 = effort 𝒂𝒓𝒎 × 𝒆𝒇𝒇𝒐𝒓𝒕

The above equation expresses the principle of moments for a lever.

Mechanical Advantage

In the equation, 𝐹₁/𝐹₂ = 𝑑₂/𝑑₁, the ratio (𝐹₁/𝐹₂) is called the mechanical advantage (𝑀𝐴) of the lever.

𝑴𝑨 = 𝑭_𝟏/𝑭_𝟐 = 𝒅_𝟐/𝒅_𝟏

We can conclude that if d₂ is larger than the d₁, then MA is greater than one. That is small effort can be used to lift a large load.

Centre of gravity

Centre of gravity (G) of a body is defined as the point where the whole weight (Gravitational force) of the body is supposed to act. While balancing a cardboard on the tip of a pencil, the tip of the pencil provides a support. The reaction of the tip is equal and opposite to 𝑀𝑔 (the total weight) of the cardboard and hence it is in translational equilibrium and it is also in rotational equilibrium. If 𝑟_𝑖 is the position vector of the 𝑖^𝑡ℎ particle of the body with respect to the centre of gravity, then the torque about centre of gravity due to force of gravity is zero. i.e 𝜏_𝑔 = ∑𝜏_𝑖 = ∑𝑟_𝑖 × 𝑚_𝑖𝑔 = 0.

As 𝑔 is same for all particles, 𝑔∑𝑚_𝑖𝑟_𝑖 = 0

∑𝑚_𝑖𝑟_𝑖 = 0 as 𝑚_𝑖 ≠ 0, 𝑟_𝑖 = 0.

Thus centre of gravity of the body coincides with the centre of mass in uniform gravity or gravity free space. This is true because the body being small, 𝑔 does not very from one point of the body to the other.

Note: If the body is so extended that 𝑔 varies from part to part of the body, then the cenre of gravity and centre of mass will not coincide. Basically centre of mass and centre of gravity are two different concepts. Centre of mass depends only on the distribution of mass of the body.

Determination of the centre of gravity of a body

(i) Consider a body of irregular shape suspend the body from the points A,B,C …… respectively.

(ii) Now make lines AA₁, BB₁, CC₁, ………..

(iii) The point where all these lines intersect is the position of the centre of gravity of the body.

Moment of Inertia (Rotational inertia)

The property of the body by virtue of which it opposes or resists changing its state of rotational motion is called rotational inertia or moment of inertia.

Explanation: Consider a rigid body of mass M consisting of n-particles. Let the body is rotating with angular velocity 𝜔 about the given axis of rotation O. Each particle of a rotating body moves in circular paths of different radii with a common centre at the axis of rotation and each particle has different linear velocity. The total kinetic energy of the body is the sum of the kinetic energies of the entire particle constituting the body.

𝐾 = (1/2)𝑚₁𝑣₁² + (1/2)𝑚₂𝑣₂² +. … … . + (1/2)𝑚_n𝑣_n²

𝐾 = (1/2)𝑚₁(r₁𝜔)² + (1/2)𝑚₂(r₂𝜔)² +. … … … + (1/2)𝑚_n(r_n𝜔)²

𝐾 = (1/2)𝜔²[𝑚₁𝑟₁² + 𝑚₂𝑟₂² +. … … + 𝑚_n𝑟_n²]

𝐾 = (1/2)𝜔²∑𝑚_𝑖𝑟_i²

𝑲 = (𝟏/𝟐)𝑰𝝎_𝟐 where 𝐼 = ∑𝑚_𝑖𝑟_𝑖²

I is called moment of Inertia. Its unit is SI system is kgm². Dimensions are 𝑀𝐿²𝑇⁰

Note: Comparing the expression for kinetic energy of a rotating body with the kinetic energy of body in linear motion, we can conclude that, the parameter moment of Inertia is the rotational analogue of mass.

This chapter will be continued in Part-2 in the next post.

Type your doubts in the comments section of this post.