CBSE NOTES CLASS 11 PHYSICS CHAPTER 7

SYSTEM OF PARTICLES AND ROTATIONAL MOTION

Rigid Body

A body with a perfectly definite and unchanging shape is called a rigid body. The distances between different pairs of particles in such a body do not change.

Centre of Mass

Centre of mass of a system is a unique point that moves as though all the mass of the system were concentrated there and all the external forces were applied at that point.

Position vector of centre of mass for n particles

If we have n particles of masses m1, m2, ...mn respectively, with position vectors, $\stackrel{\to }{{\mathrm{r}}_{1}}$, $\stackrel{\to }{{\mathrm{r}}_{2}}$,…, then the position vector $\stackrel{\to }{\mathrm{R}}$ of the centre of the mass of the system of particles is given by,

Centre of Mass of a Rigid Body

We may consider a rigid body made of very large number of small particles of mass Δm. Δm is so small that we can treat the body as continuos distribution of mass. The number of particles (atoms or molecules) in such a body is so large that it is impossible to carry out the summations over individual particles in these equations. In such cases we can take integral.

Hence we can write

= M = total mass of the body

,

&

We can thus write

The vector equation can be written as,

If we choose, the centre of mass as the origin of the coordinate system,

R (x,y,z) = 0

• The centres of mass of homogeneous (symmetrical) bodies lie at their geometric centres.

• Find centre of mass of a straight thin rod with length L and mass M. (Hint: take the origin to be geometric centre of the rod and integrate x dm from –L/2 to L/2).

• Repeat the exercise for homogeneous rings, discs, spheres and thick rods of circular or rectangular cross section.

TYPES OF MOTION

Translational Motion

In pure translational motion at any instant of time every particle of the body has the same velocity.

Example - A rectangular block sliding down an inclined plane without any sidewise movement.

Rotational Motion

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 line along which a three dimensional body rotates is termed as its axis of rotation.

Each point rotates in a circle around the axis of rotation (z-axis).

Precession

In more general cases of rotation, such as the rotation of a top or a pedestal fan, one point and not one line, of the rigid body is fixed.

The movement of the axis of the top around the vertical is termed precession.

• 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.

MOTION OF CENTRE OF MASS

For a system of particles, we have

Or M

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

Or M

$=\sum {\mathrm{m}}_{\mathrm{i}}\stackrel{\to }{{\mathrm{v}}_{\mathrm{i}}}$

Where

$\stackrel{\to }{{\mathrm{v}}_{1}}$ = $\frac{\mathrm{d}\stackrel{\to }{{\mathrm{r}}_{1}}}{\mathrm{d}\mathrm{t}}$ is the velocity of the first particle

$\stackrel{\to }{{\mathrm{v}}_{2}}$ = $\frac{\mathrm{d}\stackrel{\to }{{\mathrm{r}}_{2}}}{\mathrm{d}\mathrm{t}}$ is the velocity of the second particle etc.

and $\stackrel{\to }{\mathrm{V}}$ = $\frac{\mathrm{d}\stackrel{\to }{\mathrm{R}}}{\mathrm{d}\mathrm{t}}$ is the velocity of the centre of mass.

Differentiating with respect to time, again,

$=\sum {\mathrm{m}}_{\mathrm{i}}\stackrel{\to }{{\mathrm{a}}_{\mathrm{i}}}$

where $\stackrel{\to }{{\mathrm{a}}_{\mathrm{i}}}$ = $\frac{\mathrm{d}\stackrel{\to }{{\mathrm{v}}_{\mathrm{i}}}}{\mathrm{d}\mathrm{t}}$ is the acceleration of the ith particle and $\stackrel{\to }{\mathrm{A}}$ = $\frac{\mathrm{d}\stackrel{\to }{\mathrm{V}}}{\mathrm{d}\mathrm{t}}$ is the acceleration of the centre of mass of the system of particles.

Now, from Newton’s second law, the force acting on the ith particle is given by $\stackrel{\to }{\mathrm{F}}$i = miai.

Or M$\stackrel{\to }{\mathrm{A}}$ = $\stackrel{\to }{\mathrm{F}}$1 + $\stackrel{\to }{\mathrm{F}}$2 + ... + $\stackrel{\to }{\mathrm{F}}$n

That is, the centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point.

In the absence of net external force, the centre of mass continues to move along the same path. If one segment moves in certain direction the other segments will move such that the path of the CM is not altered.

Linear Momentum of a System of Particles

The linear momentum of a particle is defined as

$\stackrel{\to }{\mathrm{p}}$ = m $\stackrel{\to }{\mathrm{v}}$

and $\stackrel{\to }{\mathrm{F}}$ = $\frac{\mathrm{d}\stackrel{\to }{\mathrm{p}}}{\mathrm{d}\mathrm{t}}$

Let us consider a system of n particles with masses m1, m2, ..., mn respectively and velocities v1, v2, …, vn respectively.

The linear momentum of the system is defined to be the vector sum of all individual particles of the system,

$\stackrel{\to }{\mathrm{P}}$ = $\stackrel{\to }{\mathrm{p}}$1 + $\stackrel{\to }{\mathrm{p}}$2 + ... + $\stackrel{\to }{\mathrm{p}}$n

= m1 $\stackrel{\to }{\mathrm{v}}$1+m22+ ... +mn $\stackrel{\to }{\mathrm{v}}$n

$\stackrel{\to }{\mathrm{P}}=\mathrm{M}\stackrel{\to }{\mathrm{V}}$

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 its centre of mass.

Newton’s Second Law for System of Particles

Differentiating with respect to time,

Or $\frac{\mathrm{d}\stackrel{\to }{\mathrm{P}}}{\mathrm{d}\mathrm{t}}$ = $\stackrel{\to }{\mathrm{F}}$ext

This is the Newton’s second law extended to a system of particles.

Law of Conservation of Linear Momentum

If the sum of external forces acting on a system of particles is zero, then,

This is the law of conservation of the total linear momentum of a system of particles.

Or

Examples

A heavy nucleus (Ra) splits into a lighter nucleus (Rn) and an alpha particle (He). The CM of the system is in uniform motion.

If Ra were stationary, with the centre of mass at rest, the two product particles fly back to back.

A binary (double) star is a set of two stars. If there are no external forces, the centre of mass of a double star moves like a free particle,

The trajectories of the stars are a combination of (i) uniform motion in a straight line of the centre of mass and (ii) circular orbits of the stars about the centre of mass.

Separating the motion of different parts of a system into motion of the centre of mass and motion about the centre of mass is a very useful technique that helps in understanding the motion of the system.

• Revise scalar and vector products

Angular Velocity

Rate of change of angular displacement with respect to time of a body under rotational motion, is called angular velocity.

• For a rigid body all the particles of a body rotating around a fixed axis have same angular velocity.

• Each particle of the body moves in a circle which lies in a plane perpendicular to the axis and has the centre on the axis.

• The angular velocity vector $\stackrel{\to }{\mathrm{\omega }}$ is directed along the fixed axis as shown.

• The magnitude of linear velocity $\stackrel{\to }{\mathrm{v}}$ of a particle moving in a circle is related to the angular velocity of the particle $\stackrel{\to }{\mathrm{\omega }}$, by , where r is the radius of the circle (refer circular motion). It is perpendicular to both ω and r and is directed along the tangent to the circle described by the particle.

For rotation about a fixed axis, the direction of the vector $\stackrel{\to }{\mathrm{\omega }}$ does not change with time. Its magnitude may, however, change from instant to instant. For the more general rotation, both the magnitude and the direction of $\stackrel{\to }{\mathrm{\omega }}$ may change from instant to instant.

For a general case when the position vector $\stackrel{\to }{\mathrm{r}}$ = $\stackrel{\to }{\mathrm{O}\mathrm{P}}$

We have,

But as $\stackrel{\to }{\mathrm{\omega }}$ is along $\stackrel{\to }{\mathrm{O}\mathrm{C}}$

Hence, =

Angular acceleration $\stackrel{\to }{\mathrm{\alpha }}$ is defined as the time rate of change of angular velocity; Thus,

If the axis of rotation is fixed, the direction of $\stackrel{\to }{\mathrm{\omega }}$ and hence, that of $\stackrel{\to }{\mathrm{\alpha }}$ is fixed. And we can write a scalar equation

Torque on a Particle

The rotational analogue of force is moment of force. It is also referred to as torque.

If a force acts on a particle at a point P whose position with respect to the origin O is given by the position vector, the moment of the force acting on the particle with respect to the origin O is defined as the vector product

Where is the perpendicular distance of the line of action of $\stackrel{\to }{\mathrm{F}}$ form the origin and ${\mathrm{F}}_{\perp }$ (= F sin θ) is the component of $\stackrel{\to }{\mathrm{F}}$ in the direction perpendicular to $\stackrel{\to }{\mathrm{r}}$.

The moment of force (or torque) is a vector quantity.

$\stackrel{\to }{\mathrm{\tau }}$ is perpendicular to the plane containing , and its direction is given by the right handed screw rule.

• τ = 0 if r = 0, F = 0 or θ = 0o or 180o

Angular Momentum of a Particle

The angular momentum $\stackrel{\to }{\mathbit{l}}$ of the particle with respect to the origin O is defined to be

Where m = mass and = linear momentum and $\stackrel{\to }{\mathrm{r}}$ = position of the point relative

The magnitude of the angular momentum vector is

l = r p sin θ = rp = rp

where,

r is the magnitude of $\stackrel{\to }{\mathrm{r}}$,

p is the magnitude of ,

θ is the angle between $\stackrel{\to }{\mathrm{r}}$ and ,

r (= r sin θ) is the perpendicular distance of the directional line of $\stackrel{\to }{\mathrm{p}}$ from the origin

and

p(= p sin θ) is the component of in a direction perpendicular to $\stackrel{\to }{\mathrm{r}}$.

• l = 0, if p = 0, r = 0 (the particle is at origin) or if the directional line of p passes through the origin θ = 0o or 180o.

Relation between Moment of Force and Angular Momentum

We know that,

Differentiating with respect to time, we get,

Applying uv formula for multiplication, we get,

But $\stackrel{\to }{\mathrm{v}}$ = and = m, hence,

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

This is the rotational analogue of the equation $\stackrel{\to }{\mathrm{F}}=$ $\frac{\mathrm{d}\stackrel{\to }{\mathrm{p}}}{\mathrm{d}\mathrm{t}}$, which is Newton’s second law for the translational motion of a single particle.

Torque and Angular Momentum for a System of Particles

The total angular momentum of a system of particles about a given point is the vector sum of the angular momenta of individual particles. Thus, for a system of n particles,

Now, the angular momentum of the ith particle is given by

Hence

Differentiating,

But

So, total torque of the system,

Total torque $\stackrel{\to }{\mathrm{\tau }}$ is due to external and internal forces. The contribution of internal forces is zero, since the line of action of these forces is from one particle to another inside the body.

Hence,

Or

Conservation of Angular Momentum

If ${\left(\stackrel{\to }{\mathrm{\tau }}\right)}_{\mathrm{e}\mathrm{x}\mathrm{t}}$ = 0 $\frac{\mathrm{d}\stackrel{\to }{\mathrm{L}}}{\mathrm{d}\mathrm{t}}$ = 0

$\stackrel{\to }{\mathrm{L}}$ = constant

$\stackrel{\to }{\mathrm{L}}$ can be represented as components along x, y and z directions and all three components of angular momentum are conserved. The above equation can therefore be written as,

Lx = k1, Ly = k2, Lz = k3

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, or equivalently, the body has neither linear acceleration nor angular acceleration. This means

(i) The total force, i.e. the vector sum of the forces, on the rigid body is zero;

$\stackrel{\to }{\mathrm{F}}$1 + $\stackrel{\to }{\mathrm{F}}$2 + … + $\stackrel{\to }{\mathrm{F}}$n = Σ $\stackrel{\to }{\mathrm{F}}$i = 0

Or alternatively,

ΣFix = ΣFiy = ΣFiz

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 the translational equilibrium of the body.

(ii) The total torque, i.e. the vector sum of the torques on the rigid body is zero,

$\stackrel{\to }{\mathrm{\tau }}$1 + $\stackrel{\to }{\mathrm{\tau }}$2 + … + $\stackrel{\to }{\mathrm{\tau }}$n = Σ $\stackrel{\to }{\mathrm{\tau }}$i = 0

Or alternatively,

Σ τix = Σ τiy = Σ τiz = 0

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 the rotational equilibrium of the body.

Moment of Couple

A pair of equal and opposite forces with different lines of action is known as a couple. A couple produces rotation without translation.

• For example, our fingers apply a couple to turn the lid.

The moment of the couple

Lever

An ideal lever is essentially a light (i.e. of negligible mass) rod pivoted at a point along its length. This point is called the fulcrum.

A seesaw on the children’s playground is a typical example of a lever.

Two forces F1 and F2, parallel to each other and usually perpendicular to the lever act on the lever at distances d1 and d2 respectively from the fulcrum.

The lever is a system in mechanical equilibrium. Let $\stackrel{\to }{\mathrm{R}}$ be the reaction of the support at the fulcrum; $\stackrel{\to }{\mathrm{R}}$ is directed opposite to the forces F1 and F2. For translational equilibrium,

R F1F2 = 0

For considering rotational equilibrium, the sum of moments about the fulcrum must be zero, i.e.,

d1 F1 – d2 F2 = 0

or d1 F1 = d2 F2

In the case of the lever force F1 is some weight to be lifted. It is called the load and its distance from the fulcrum d1 is called the load arm.

Force F2 is the effort applied to lift the load; distance d2 of the effort from the fulcrum is the effort arm.

Mechanical Advantage (M.A.) = = $\frac{{\mathrm{d}}_{2}}{{\mathrm{d}}_{1}}$

Centre of Gravity

The CG of a body may be defined as that point where the total gravitational torque on the body is zero.

If $\stackrel{\to }{\mathrm{r}}$i is the position vector of the ith particle of an extended body with respect to its CG, then,

$\stackrel{\to }{\mathrm{\tau }}$g = Σ $\stackrel{\to }{\mathrm{\tau }}$i = Σ ${\mathrm{g}}_{\mathrm{i}}{\mathrm{m}}_{\mathrm{i}}\stackrel{\to }{\mathrm{r}}$i = 0

If g is constant throughout the body CM and CG will coincide. In case g varies from from part to part of the body, then the centre of gravity and centre of mass will not coincide.

Finding Centre of Gravity of a Body

If we suspend the body from some point like A, the vertical line through A (AA1) passes through the CG. We then suspend the body through other points like B and C. The intersection of the verticals gives the CG.

Moment of Inertia (I)

Moment of inertia of a body about a given axis is the sum of the products of masses and squares of respective perpendicular distances from the axis of rotation, for all the particles of the body.

$\mathrm{I}={\sum }_{\mathrm{i}=0}^{\mathrm{n}}{\mathrm{m}}_{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^{2}$

SI unit of moment of inertia is kg m2.

The moment of inertia is independent of the magnitude of the angular velocity. It is a characteristic of the rigid body and the axis about which it rotates.

The moment of inertia of a rigid body depends on the mass of the body, its shape and size; distribution of mass about the axis of rotation, and the position and orientation of the axis of rotation.

Moment of inertia is the rotational analogue of mass.

Relation of KE and I

For a body rotating about a fixed axis, each particle of the body moves in a circle with linear velocity given by v = r ω.

Let the body be made of n particles with masses mi, linear velocity vi, position vectors ri and angular velocity ωi.

vi = ri ωi

Then the KE of the ith particle will be,

ki = $\frac{1}{2}$ mi vi2 = $\frac{1}{2}$ mi ri2 ωi2

and KE for the whole body will be,

K = Σ ki = $\frac{1}{2}$ Σ mi ri2 ωi2

Since angular velocity is same for all particles, we can write,

K = Σ ki = $\frac{1}{2}$ ω2 Σ mi ri2 = $\frac{1}{2}$ I ω2

If ω = 1, then I = 2 KE

The radius of gyration of a body about an axis may be defined as the distance from the axis of a point mass whose mass is equal to the mass of the whole body and whose moment of inertia is equal to the moment of inertia of the body about the axis.

We know,

I = Σ mi ri2 = M Σ $\frac{{{\mathrm{r}}_{\mathrm{i}}}^{2}}{\mathrm{n}}$ = M k2

Where k =

The radius of gyration of rigid body depends on the mass of the body, its shape and size; distribution of mass about the axis of rotation, and the position and orientation of the axis of rotation.

Calculations of moment of inertia for some regular shaped bodies

(a) A thin ring of radius R and mass M, rotating in its own plane around its centre with angular velocity ω.

Each mass element of the ring is at a distance R from the axis, and moves with a speed Rω. The moment of inertia,

The kinetic energy therefore, is

K = $\frac{1}{2}$ I ω2 = $\frac{1}{2}$ MR2 ω2

(b) A rigid massless rod of length l with a pair of small masses, rotating about an axis through the centre of mass perpendicular to the rod. Each mass $\frac{\mathrm{M}}{2}$ is at a distance $\frac{\mathrm{l}}{2}$ from the axis.

The moment of inertia of the masses is therefore given by

(c) Moment of inertia of uniform rod about a perpendicular bisector

Consider a homogeneous and uniform rod of mass M and length L.

Consider middle point O to be the origin of the rod. Also consider an element of the rod between the distance x and x+dx from the origin. Since the rod is uniform so its density

Hence mass of the element

Perpendicular distance of this element from line XY is x, so that moment of inertia of this element about XY is

$=\mathrm{M}/\mathrm{L}{\left[\frac{{\mathrm{x}}^{3}}{3}\right]}_{-\mathrm{L}/2}^{\mathrm{L}/2}=\frac{\mathrm{M}{\mathrm{L}}^{2}}{12}$

(d) Moment of inertia of the uniform circular plate about its axis

To find the moment of inertia of the disk with radius R and mass M about the axis OX, draw two concentric circles of radii x and x+dx having their common centres at O, so that they form a ring

Area of the ring =2πx dx

Mass of the ring would be

Moment of inertia of this ring about axis YY’ would be

Moment of inertia of complete disc

$=\frac{2\mathrm{M}}{{\mathrm{R}}^{2}}{\left[\frac{{\mathrm{x}}^{4}}{4}\right]}_{0}^{\mathrm{R}}$

$=\frac{\mathrm{M}{\mathrm{R}}^{2}}{2}$

(e) Moment of inertia of a uniform sphere of radius R about the axis through its center

Consider a sphere of mass M and radius R. Let us divide this sphere into thin discs as shown in the figure

The radius of the disc will be

Volume of theis disc

dV = π r2dx = π dx

Masss of this disc dm = ρ dV

Moment of inertia of this disc would be,

dI = dm r2 = dx

Moment of inertia of the sphere

Now the mass of the sphere,

$⇒\mathrm{\rho }=\frac{3\mathrm{M}}{4\mathrm{\pi }{\mathrm{R}}^{3}}$

Putting the value of ρ, we get,

$\mathrm{I}=\frac{2}{5}\mathrm{M}{\mathrm{R}}^{2}$

Flywheel

A flywheel is a mechanical device specifically designed to efficiently store rotational energy. Because of its large moment of inertia, the flywheel resists the sudden increase or decrease of the speed of the vehicle. It allows a gradual change in the speed and prevents jerky motions, thereby ensuring a smooth ride for the passengers on the vehicle.

Theorem of perpendicular axes

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.

Moment of inertia of the body around z-axis = moment of inertia around x-axis + moment of inertia around y-axis, or,

Iz = Ix + Iy

Proof:

Let, the body consists of n number of particles of masses m1, m2, m3, mn and their positions vectors are $\stackrel{\to }{\mathrm{r}}$1, $\stackrel{\to }{\mathrm{r}}$2, ……$\stackrel{\to }{\mathrm{r}}$n respectively.

Let m1 be located at point P (x1, y1).

So, r12 = x12 +y12

The moment of inertia of the particle of mass m1, about OZ axis is,

I1 = m1r12 = m1 (x12 + y12)

Similarly the of inertia of the particle of mass m2 about OZ axes is,

I2 = m2 (x22 + y22)

Therefore, the moment of inertia of the whole body about OZ axis is,

I= I1 + I2 + … …+ In

= m1(x12 + y12)+m2(x22+y22)+…+mn(xn2 + yn2)

= (m1x12+m2x22 +…..+ mnxn2) + (m1y12+m2y22 +…..+ mnyn2)

Therefore, Iz = Ix + Iy

• Find MI of a ring about one of its diameters.

Theorem of parallel axes

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 the given figure,

Iz′ = Iz + Ma2

Proof:

Let the ith particle have mass mi and distance from the axis of rotation (Z) be ri. Le Z’ be another axis parallel to Z-axis and at a distance ‘a’ from it.

Then,

Iz’ = Σ mi (a+ri)2

= Σ mi (a2+2ari+ri2)

= Σ mi ri2 + Σ mi a2 + Σ mi 2ari

Since ri will have +ve as well as negative values, the sum of moments of weight of all the particles of the body will be zero, i.e.,

Σ mi ri = 0

Hence the third term in the summation will be zero.

Or Iz’ = Iz + Ma2

• Find MI of a ring about one of its tangents.

• Find MI of rod about an axis perpendicular to the rod and passing through one of the ends.
 Moments of Inertia of some regular shaped bodies about specific axes SN Body Type Axis MI 1 Thin circular ring, radius R Perpendicular to plane, at centre MR2 2 Thin circular ring, radius R Diameter $\frac{\mathrm{M}{\mathrm{R}}^{2}}{2}$ 3 Thin rod, length L Perpendicular to rod, at mid point $\frac{\mathrm{M}{\mathrm{R}}^{2}}{12}$ 4 Circular disc, radius R Perpendicular to disc at centre $\frac{\mathrm{M}{\mathrm{R}}^{2}}{2}$ 5 Circular disc, radius R Diameter $\frac{\mathrm{M}{\mathrm{R}}^{2}}{4}$ 6 Hollow cylinder, radius R Axis of cylinder MR2 7 Solid cylinder, radius R Axis of cylinder $\frac{\mathrm{M}{\mathrm{R}}^{2}}{2}$ 8 Solid sphere, radius R Diameter $\frac{2}{5}\mathrm{M}{\mathrm{R}}^{2}$

Torque and Moment of Inertia (τ = I α)

Consider a rigid body rotating about a given axis with a uniform angular acceleration α under action of a torqueτ. Let the body be divided into n particles of masses m1, m2, …, mn with perpendicular distance r1, r2, …, rn respectively from the fixed axis of rotation. Since the body is rigid, angular acceleration of all the particles is same. However the linear accelerations (ai) are different due to different distances of particles from the centre of circles on the axis.

Then

a1 = r1 α,

a2 = r2 α, ….,

an = rn α

And forces

f1=m1 a1,

f2=m2 a2, ….,

fn=mn an

Therefore,

f1=m1 r1 α,

f2=m2 r2 α, ….,

fn=mn rn α

Moments of these forces around the fixed axis,

τi = ri fi = mi ri2 α

Total torqueτ = Σ mi ri2 α = α Σ mi ri2 = I α

Moment of inertia of a body is numerically equal to the torque acting on the body rotating with unit angular acceleration about it.

Angular Momentum and Moment of Inertia

Since the body is rigid, angular velocity (ω) of all the particles is same. However the linear velocities are different due to different distances of particles from the axis.

Now $\stackrel{\to }{\mathrm{v}}$i = $\stackrel{\to }{\mathrm{r}}$i $×\stackrel{\to }{\mathrm{\omega }}$

Linear momentum of particles

$\stackrel{\to }{\mathrm{p}}$i= mi $\stackrel{\to }{\mathrm{v}}$i = mi $\stackrel{\to }{\mathrm{r}}$i $×\stackrel{\to }{\mathrm{\omega }}$

Angular momentum = $\stackrel{\to }{\mathrm{l}}$i = $\stackrel{\to }{\mathrm{r}}$i $×\stackrel{\to }{\mathrm{p}}$i

$\mathrm{l}$i = mi ri2 ω

Total angular momentum of the body about given axis

L = Σ mi ri2 ω = I ω

(This relation is similar to p = mv)

DYNAMICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS

Since the axis of the rotating body is fixed, only those components of torques, which are along the direction of the fixed axis, need to be considered. Only these components can cause the body to rotate about the axis. A component of the torque perpendicular to the axis of rotation will tend to turn the axis from its position. The perpendicular components of the torques, therefore, need not be taken into account. This means that for our calculation of torques on a rigid body:

1. We need to consider only those forces that lie in planes perpendicular to the axis.

Forces which are parallel to the axis will give torques perpendicular to the axis and need not be taken into account.

2. We need to consider only those components of the position vectors which are perpendicular to the axis. Components of position vectors along the axis will result in torques perpendicular to the axis and need not be taken into account.

Work done by a torque

Let us consider a rigid body rotating about a fixed axis, which is taken as the z-axis (perpendicular to the plane of the page in the figure)

Now we need to consider only those forces which lie in planes perpendicular to the axis. Let F1 be one such force acting on a particle of the body at point P with its line of action in a plane perpendicular to the axis.

The particle at P describes a circular path of radius r1 with centre C on the axis; CP1 = r1.

In time Δt, the point moves to the position P′.

The displacement of the particle has a magnitude ds1 = r1dθ and direction tangential at P to the circular path.

Here dθ is the angular displacement of the particle, dθ = ∠PCP′.

The work done by the force on the particle is,

dW1 = $\stackrel{\to }{\mathrm{F}}$1. d$\stackrel{\to }{\mathrm{s}}$1

= F1 ds1 cos Φ1

= F1(r1 dθ) sin α1

where Φ1 is the angle between F1 and the tangent at P, and α1 is the angle between F1 and the radius vector $\stackrel{\to }{\mathrm{O}\mathrm{P}}$

Φ1 + α1 = 90°.

The torque due to about the origin is $\stackrel{\to }{\mathrm{O}\mathrm{P}}$ × $\stackrel{\to }{\mathrm{F}}$1.

Now $\stackrel{\to }{\mathrm{O}\mathrm{P}}$ = $\stackrel{\to }{\mathrm{O}\mathrm{C}}$ + $\stackrel{\to }{\mathrm{C}\mathrm{P}}$.

Since $\stackrel{\to }{\mathrm{O}\mathrm{C}}$ is along the axis, the torque resulting from it need not be considered.

The effective torque due to $\stackrel{\to }{\mathrm{F}}$1 is,

$\stackrel{\to }{\mathrm{\tau }}$1= $\stackrel{\to }{\mathrm{C}\mathrm{P}}$ × $\stackrel{\to }{\mathrm{F}}$1;

And it is directed along the axis of rotation and has a magnitude

τ1= r1F1 sin α1,

Therefore, dW1 = τ1

If there are more than one forces acting on the body, the work done by all of them can be added to give the total work done on the body.

Denoting the magnitudes of the torques due to the different forces as τ1, τ2, … etc,

dW = (τ12 + ...) dθ

Since all the torques considered are parallel to the fixed axis, the magnitude τ of the total torque is the algebraic sum of the magnitudes of the torques, i.e.,

τ = τ1 + τ2 + .....

Therefore, dW = τ dθ

This is similar to expression

dW = F ds

for linear (translational) motion.

Deriving the equation of torque starting with equation of KE

We know that KE = $\frac{1}{2}$2

Hence power = rate of change of kinetic energy

Equating both the values of power we get,

τω = $\mathrm{I}\mathrm{\omega }\mathrm{\alpha }$

⇒ τ = $\mathrm{I}\mathrm{\alpha }$

This similar to Newton’s second law for linear motion

F = ma

Angular Momentum

Since the body is rigid, angular velocity (ω) of all the particles is same. However the linear velocities are different due to different distances of particles from the axis.

Now $\stackrel{\to }{\mathrm{v}}$i = $\stackrel{\to }{\mathrm{r}}$i $×\stackrel{\to }{\mathrm{\omega }}$

Linear momentum of particles

$\stackrel{\to }{\mathrm{p}}$i= mi $\stackrel{\to }{\mathrm{v}}$i = mi $\stackrel{\to }{\mathrm{r}}$i $×\stackrel{\to }{\mathrm{\omega }}$

Angular momentum = $\stackrel{\to }{\mathrm{l}}$i = $\stackrel{\to }{\mathrm{r}}$i $×\stackrel{\to }{\mathrm{p}}$i

But $\stackrel{\to }{\mathrm{r}}$i = $\stackrel{\to }{\mathrm{O}\mathrm{P}}$ = $\stackrel{\to }{\mathrm{O}\mathrm{C}}$ + $\stackrel{\to }{\mathrm{C}\mathrm{P}}$.

$\stackrel{\to }{\mathrm{l}}$i = $\stackrel{\to }{\left(\mathrm{O}\mathrm{C}}$ + $\stackrel{\to }{\mathrm{C}\mathrm{P}}$i $\right)×\stackrel{\to }{\mathrm{p}}$i

= $\stackrel{\to }{\mathrm{O}\mathrm{C}}×\mathrm{m}\stackrel{\to }{\mathrm{v}}$i + $\stackrel{\to }{\mathrm{C}\mathrm{P}}$i $×\mathrm{m}\stackrel{\to }{\mathrm{v}}$i

Now we can write $\stackrel{\to }{\mathrm{C}\mathrm{P}}$i $×\mathrm{m}\stackrel{\to }{\mathrm{v}}$i = ri$\stackrel{̂}{\mathrm{k}}$ (parallel to axis of rotation)

Or ${\stackrel{\to }{\mathrm{l}}}_{\mathrm{i}\mathrm{z}}$= ri$\stackrel{̂}{\mathrm{k}}$

Total angular momentum therefore is

$\stackrel{\to }{\mathrm{L}}$ = $\stackrel{\to }{\mathrm{O}\mathrm{C}}×\mathrm{m}\stackrel{\to }{\mathrm{v}}$i = ${\stackrel{\to }{\mathrm{L}}}_{\mathrm{z}}+{\stackrel{\to }{\mathrm{L}}}_{\perp }$

Since only those components of torque are to be considered which are along the axis, we have,

But

Hence we get

• Angular Momentum is conserved. When moment of inertia increases, 𝛚 decreases and vice versa.

• For example, skaters and classical dancers performing a pirouette on the toes of one foot, bring their hands closer to the body to increase angular speed.

• A circus acrobat and a diver also take advantage of this principle.

Rolling Motion

Rolling motion is a combination of rotation and translation.

Consider a disc rolling without slipping on a level surface. This means that at any instant of time the bottom of the disc which is in contact with the surface is at rest on the surface.

The translational motion of a system of particles is the motion of its centre of mass.

Let $\stackrel{\to }{\mathrm{v}}$cm be the velocity of the centre of massand therefore the translational velocity of the disc.

Since the centre of mass of the rolling disc is at its geometric centre C, $\stackrel{\to }{\mathrm{v}}$cm is the velocity of C. It is parallel to the level surface.

The rotational motion of the disc is about its symmetry axis, which passes through C and is perpendicular to the plane of the disc. Thus, the velocity of any point of the disc consists of two parts, one is the translational velocity $\stackrel{\to }{\mathrm{v}}$cm and the other is the linear velocity $\stackrel{\to }{\mathrm{v}}$r on account of rotation.

The magnitude of $\stackrel{\to }{\mathrm{v}}$r is vr = r ω, where ω is the angular velocity of the rotation of the disc about the axis and r is the distance of the point from the axis (i.e. from C).

The velocity $\stackrel{\to }{\mathrm{v}}$r is directed perpendicular to the radius vector of the given point with respect to C.

Let us consider the velocity of the point P2 ($\stackrel{\to }{\mathrm{v}}$2)

$\stackrel{\to }{\mathrm{v}}$2 = $\stackrel{\to }{\mathrm{v}}$r + $\stackrel{\to }{\mathrm{v}}$cm

$\stackrel{\to }{\mathrm{v}}$r here is perpendicular to CP2 .

$\stackrel{\to }{\mathrm{v}}$2 is perpendicular to the line PoP2. Therefore the line passing through Po and parallel to ω is called the instantaneous axis of rotation.

Now |$\stackrel{\to }{\mathrm{v}}$r|= rω,

• At point Po, $\stackrel{\to }{\mathrm{v}}$R and $\stackrel{\to }{\mathrm{v}}$cm are directed opposite to each other and the velocity of the point at that instant is zero , hence,

|$\stackrel{\to }{\mathrm{v}}$R| = $\stackrel{\to }{\mathrm{v}}$cm = Rω

• Velocity at the top $\stackrel{\to }{\mathrm{v}}$1 = $\stackrel{\to }{\mathrm{v}}$R + $\stackrel{\to }{\mathrm{v}}$cm = 2Rω and is directed parallel to the level surface.

Kinetic Energy of Rolling Motion

Total KE = Translational KE + Rotational KE

K = KT + Kr

= $\frac{1}{2}$ m vcm2 + $\frac{1}{2}$ I ω2

For a rolling cylinder,

Motion of a cylinder rolling without slipping on an inclined plane

Consider a round object (this could be a cylinder, hoop, sphere or spherical shell) having mass M, radius R and rotational inertia I about its center of mass, rolling without slipping down an inclined plane.

The force of gravity, Mg, acting straight down is resolved into components parallel and perpendicular to the incline

Since there is no motion perpendicular to the inclined plane,

R = Mg cos θ

Since the object rolls without slipping there is a force of friction f, acting on the object, at the point of contact with the incline, in the direction up the incline

Force of friction = f = μR = μMg cos θ

If the linear acceleration down the plane is ‘a’, then,

Ma = Mg sin θ - f

Consider the centre of mass of the cylinder to be the origin. All the forces except f pass through the centre of mass of the cylinder, so only f will cause a torque,

τ = r × f = r μMg cos θ

But we have τ = Iα = $\frac{\mathrm{I}\mathrm{a}}{\mathrm{r}}$

Hence r f = $\frac{\mathrm{I}\mathrm{a}}{\mathrm{r}}$f = $\frac{\mathrm{I}\mathrm{a}}{{\mathrm{r}}^{2}}$

So, Ma = Mg sin θ - $\frac{\mathrm{I}\mathrm{a}}{{\mathrm{r}}^{2}}$

Or a (M +) = Mg sin θ

Now, I = Mk2

Now moment inertia of a of cylinder about the axis of symmetry,

I = $\frac{1}{2}$ Mr2 or ${\mathrm{k}}^{2}$ = $\frac{1}{2}$ r2

We can see that a < g

Now frictional force f = $\frac{\mathrm{I}\mathrm{a}}{{\mathrm{r}}^{2}}$

Putting the value of ‘a’ and I, we get

f = Mr2 × × $\frac{1}{{\mathrm{r}}^{2}}$ = $\frac{1}{3}$Mg sin θ

So, f < Mg

Now f = μ R or μ = f/R

Laws of Rotational Motion

1. A body continues to be in a state of rest or in a state of uniform rotation about a given axis unless an external torque is applied on the body.

2. Rate of change of angular momentum of a body about a given axis is directly proportional to the external torque applied on the body.

3. When a rigid body A exerts a torque on another rigid body B in contact with it, then the body B would exert and equal and opposite torque on body A.

Formulae for Translational and Rotational Motion

 S.No. Linear Motion Rotational Motion 1 Displacement s Angular displacement θ 2 Velocity v = $\frac{\mathrm{d}\mathrm{s}}{\mathrm{d}\mathrm{t}}$ Angular velocity ω = $\frac{\mathrm{d}\mathrm{\theta }}{\mathrm{d}\mathrm{t}}$ 3 Acceleration a = $\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{t}}$ Angular acceleration α = $\frac{\mathrm{d}\mathrm{\omega }}{\mathrm{d}\mathrm{t}}$ 4 v = u + at ω = ωo + αt 5 s = ut + $\frac{1}{2}$ at2 θ = ωot + $\frac{1}{2}$ αt2 6 v2 – u2 = 2as ω 2 – ωo 2 = 2αθ 5 Mass M Moment of inertia I 6 Force F = Ma Torque τ = I α 7 Work dW = F ds Work dW = τ dθ 8 Kinetic energy K = $\frac{1}{2}$Mv2 Kinetic energy K = $\frac{1}{2}$ Iω2 9 Power P = F v Power P = τω 10 Linear momentum p = Mv Angular momentum L = Iω