system of particles and rotational motion

System of particles And rotational motion

-By Anshul Sharma

content 7.1 Introduction 7.2 Centre of mass 7.3 Motion of centre of mass 7.4 Linear momentum of a system of particles 7.5 Angular momentum and Angular velocity 7.6 Torque and angular momentum 7.7 Equilibrium of a rigid body 7.8 Moment of inertia 7.9 Theorems of perpendicular and parallel axes 7.10 Kinematics of rotational motion about a fixed axis 7.11 Comparison of Translational and Rotational Motion 7.12 Angular momentum in case of rotation about a fixed axis 7.13 Rolling motion

7.1 INTRODUCTION

An extended body, in the first place, is a system of particles. We shall begin with the consideration of motion of the system as a whole. The centre of mass of a system of particles will be a key concept here. We shall discuss the motion of the centre of mass of a system of particles and usefulness of this concept in understanding the motion of extended bodies. A large class of problems with extended bodies can be solved by considering them to be rigid bodies. Ideally a rigid body is a body with a perfectly definite and unchanging shape. The distances between different pairs of such a body do not change. It is evident from this definition of a rigid body that

7.1.1 What kind of motion can a rigid body have? Translational motion

Rotational motion

Translation + Rotational motion

7.2 CENTRE OF MASS We may define centre of mass of a body or a system of bodies as a point at which the entire mass of the body/system of bodies, is supposed to be concentrated.

7.2.1Centre of Mass of a Two Particle System Consider a system of two particles of mass m1 and m2 located at A and B respectively, where

Let C be the position of centre of mass of the system of two particles. It would lie on the line joining A and B. Let be the position  vector of centre of mass. r,

  v1 evaluate , v2 To

suppose

are velocities of particles m1 and m2

Respectively at any instant t. Then  dr 1  dr 2 v1 =  dt f1

and

v2 =

dt

...(1)



Letf 2 = external force applied on particle of mass m1  F12

= external force applied on  particle of mass m2 F 21

m2

= internal force on m1 due to

  p1 = m1v1

...( 2)

= internal force on m2 due to m1 Linear momentum of particle m1 is

which is    dp1 ∴ = f1 + F12 dt    d u sin g ( 2), ( m1 v1 ) = f1 + F12 dt

...( 3)

Similarly, the equation of motion of second   as  be written d particle may ( m v ) = f +F ...( 4) dt

2

2

2

21

   and d d (4), we get Adding (3) ( m1 v1 ) + ( m2 v2 ) = f1 + F12 dt As

∴

 F12

dt  = − F21

  d [( m1 v1 ) +m2 v2 ] dt

  + f 2 + F21

   = f1 + f 2 = f

...( 5)

 f 1 = total external force on the system

Where of two particles.

Using (1),we get

  dr2   d  dr1 m1 + m2 = f   dt  dt dt      d d or (m1 r1 + m2 r2 ) = f  dt  dt  or

    d2 d (m1 r1 + m2 r2 ) = f 2  dt  dt 

Multiplying numerator and denominator of left

   d ( m1 r1 + m2 r2 ) ( m1 + m2 ) 2 = f ( m1 + m2 ) dt   ( m1 r1 + m2 r2 )  Let us put =r ( m1 + m2 )  d2  ∴ ( m1 + m2 ) 2 (r ) = f dt 2

   m1 +we m2get )r = m1 r1 + m2 r2 From ((7),

...( 6) ...( 7) ...( 8)

...(9)

Hence position vector of centre of mass of a two particle system is such that the product of total mass of the system and position vector of centre of mass is equal to sum of the products of masses of the two particles and their respective

7.2.2Centre of Mass of a System of N Particles

    m1 r1 + m2 r2 + ...mn rn r= m1 + m2 + ...mn

7.3 MOTION OF Equipped with the definition of the centre of mass, we are now in a position to discuss its CENTRE OF MASS physical importance for a system of particles. we know :

second particle etc. and is the velocity of the centre of mass. Note that we assumed the masses m1, m2, ... etc. do not change in time. We have therefore, treated them as constants in differentiating the equations with respect to time. Differentiating Eq.(7.8) with respect to time, we obtain

Now, from Newton’s second law, the force acting on the first particle is given by . The force acting on the second particle is given by on. Eq. (7.9) may be written as

Thus, the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.

(7.11)

7.4 LINEAR MOMENTUM OF A SYSTEM OF PARTICLES The linear momentum of an object is defined as the motion contained in a body OR the product of the mass and velocity is known as momentum.

Momentum is a vector quantity and its direction is the same as that for velocity; And it has dimension ML/T. In SI system, the momentum has the units kg m/s. the Newton’s second law can be written as:

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. Differentiating Eq. (7.15) with respect to time,

This is the statement of Newton’s second law extended to a system of particles. Suppose now, that the sum of external forces acting on a system of particles is zero. Then from Eq.(7.17)

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 total linear momentum of a system of particles.

7.5 Angular Momentum and Angular Velocity In general, each component of the total

 L

 ω   l ω r’

angular momentum depends on all the components of the angular velocity.

(

        2 r × m ( ω × r ) = m r ω − ( r L = ∑ ri × m i v i = ∑ i ∑ i i i i i ω) ri i

i

i

L x =  − ∑ mi z i x i ω ;  i 

(

)

L y =  − ∑ mi zi yi ω ;  i 

(

L z = ∑ mi ri2ω − z i ω ⋅ zi =  ∑ mi ri2 − zi2 i i

) ω =  ∑ m r' i

2 i i ω



)

7.6 Torque and angular Torque , about the reference point O, due to a momentum force F exerted on a particle, is defined as the vector product of the position relative to the reference point and force

In other words Turning effect of force is known as torque .  r O

 τ

The angular momentum of a particle, about the reference point O, is defined as the vector product of the position, relative to the reference point, and momentum of the particle In other words motion contained in a body while it is moving in circular part is known as angular momentum.  l  r O

 p

7.7 Equilibrium of a rigid body F1

O

A

F 3

F2

A rigid object is in equilibrium, if and only if the following conditions are satisfied: (a) the net external force is a zero vector;

7.7.1 couple A pair of equal and opposite forces with different lines of action is known as a couple. A couple produces rotation without translation. When we open the lid of a bottle by turning it, our fingers are applying a couple to the lid.

Another known example is a compass needle in the earth’s magnetic field. The earth’s magnetic field exerts equal forces on the north and south poles. The force on the North Pole is towards the north, and the force on the South Pole is toward the south. Except when the needle points in the north-south direction; the two forces do not have the same line of action. Thus there is a couple acting on the needle due to the earth’s magnetic field.

7.7.2 Principle of moments An ideal lever is essentially a light rod pivoted at a point along its length. This point is called the fulcrum. A see saw 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, as shown here, act on the lever at distances d1 and d2 respectively from the fulcrum.

The lever is a system in mechanical equilibrium. Let R be the reaction of the support at the fulcrum; R is directed opposite to the forces F1 and F2. For translational equilibrium, For considering rotational equilibrium we take the moments about the fulcrum; the sum of moments must be zero, Normally the anticlockwise (clockwise) moments are taken to be positive (negative). Note R acts at the fulcrum itself and has zero moment about the fulcrum.

In the case of the lever force F1 is usually 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. Eq. (ii) can be written as or load arm × load = effort arm × effort The above equation expresses the principle of moments for a lever. Incidentally the ratio is called the Mechanical Advantage (M.A.);

If the effort arm d2 is larger than the load arm, the mechanical advantage is greater than one. Mechanical advantage greater than one means

7.7.3 Centre of gravity Centre of gravity of the body is defined as the point where the whole weight of the body were supposed to act. The CG of the cardboard is so located that the total torque on it due to the forces …. etc. is zero. If is the position vector of the particle of an extended body with respect to its CG, then the torque about the CG, due to the force of gravity on the particle is . The total gravitational torque about the CG is zero, i.e. We notice that in above Eq. , g is the same for all particles, and hence it comes out of the summation. This gives, since g is nonzero,

Example of Center of gravity       τ = ∑ ri × mi g =  ∑ miri  × g = i  i     = Mrcm × g = rcm × Mg

 ri  Wi

The center of gravity in a uniform gravitational field is at the center of mass.

Note: Not applicable to a nonuniform gravitational field

gravitational torque

7.8 Moment of inertia Moment of inertia or rotational inertia of a body about a given axis of rotation is defined as the sum of the product of the masses of the particles and square of the distances from the axis of rotation. It is denoted by I and given by;

We now apply the definition Eq. (7.34), to calculate the moment of inertia in two simple

Moment of inertia of a uniform thin rod

about an end L

Iy = ∫ x2

y

0

dx

3 L

M M x dx = ⋅ L L 3

0

1 = ML2 3

x L

about the center L/2

Icm = ∫ x 2 −L / 2

3 L/2

M x M dx = ⋅ L 3 L

−L / 2

1 ML2 = 12

Moment of inertia of a uniform circle r 

dr d

R2 M 2 2 r ⋅ rdϕdr = I A = ∫ r dm = ∫ ∫ 2 πR circle 00

M R 3  2π  1 M R4 2 = r d ϕ dr = MR 2 π ⋅ ⋅ =   ∫ ∫ 2 πR 2 0  0  πR 2 4

Moment of Inertia for system of particles and a continuous body A

system of particles:

IA =

2 m r ' ∑ i i i

r’ ri’ dm mi

continuous body A

Moments of Inertia of some regular shaped bodies about specific axes

7.8.1 radius of gyration Distance of a point in a body from the axis of rotation, at which if whole of the mass of the body were supposed to be concentrated, its moment of inertia about the axis of rotation will be same as that determined by actual distribution of mass of the body is called radius of gyration.

it is denoted by K. Notice from the Table 7.1 that in all cases, we can write , where k has the dimension of length. For a rod, about the perpendicular axis at its midpoint,

7.9 THEOREMS OF PERPENDICULAR AND PARALLEL AXES

These are two useful theorems relating to moment of inertia. We shall first discuss the theorem of perpendicular axes and its simple yet instructive application in working out the moments of inertia of some regular-shaped bodies.

7.9.1 Theorem of axes It perpendicular 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.

The figure shows a planar body. An axis perpendicular to the body through a point O is taken as the z-axis. Two mutually perpendicular axes lying in the plane of the body and concurrent with z-axis, i.e. passing through O, are taken as the x and y-axes. The theorem states that;

Derivation: Let us consider a plane lamina lying in the XOY plane. The lamina is made up of a large number of

Now PNl= x, PN = y Moment of inertia about X-axis = my2 Moment of inertia of the whole of lamina about X-axis Moment of inertia of the whole of lamina about Y-axis

Moment of inertia of the whole of lamina about Z-axis

7.9.2 Theorem 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.

As shown in the Figure, z and z′ are two parallel axes separated by a distance a. The z-axis passes through the centre of mass O of the rigid body. Then according to the theorem of parallel axes;

where and are the moments of inertia of the body about the z and z′ axes respectively, M is the total mass of the body and a is the perpendicular distance between the two parallel axes.

Derivation: The Parallel Axis Theorem (from first year physics) tells us that the moment of inertia is the sum of the moment of inertia about the mass center (Ic) plus the product of the mass (M) and the square of the distance (d) from the axis of rotation to the mass center.   

The same holds for the unknown moments of inertia , which is about a parallel axis that is a distance d + x from the mass center:  I' = Ic + M(d + x)2 Now substitute what we've already found is equal to Ic: I' = (I - Md2) + M(d + x)2   = (I - Md2) + M(d2 + 2dx + x2)

The kinematical quantities in rotational motion, angular displacement (θ), angular velocity (ω) and angular acceleration (α) respectively correspond to kinematic quantities in linear motion, displacement (x), velocity (v) and acceleration (a). We know the kinematical equations of linear motion with uniform (i.e. constant) acceleration:

7.10 Kinematics of rotational motion about a fixed axis

where initial displacement and initial velocity. The word ‘initial’ refers to values of the quantities at t = 0. The corresponding kinematic equations for rotational motion with uniform angular acceleration are:

where initial angular displacement of the rotating body, and initial angular velocity of the body.

7.11Comparison of Translational and Rotational Motion

7.13 rolling motion In order to bring out characterizing aspects of rolling motion, we consider a disk, which is rolling without sliding (simply referred as rolling) smoothly on a horizontal surface such that its center of mass translates with a velocity "vC" in x-direction. Rolling without sliding: “Rolling without sliding” means that the point on the rim in contact with the surface changes continuously as the disk rotates while translating ahead. If point "A" is in contact at a given time "t", then another neighboring point "B" takes up the position immediately after, say, at a time instant, t+dt. In case the disk slides while

The terms “Rolling without sliding”, "pure rolling" or simply "rolling" refer to same composite motion along a straight line.

7.13.1Rolling motion as a combined motion: Rolling is considered as the combination of pure rotation and pure translation.

Pure rotational For pure rotation, we consider that the rotating disk rotates about a fixed axis with angular velocity, "ω" such that : ⇒ ω = vC /R

⇒ v = vC = ωR However, if we consider a particle inside the disk at radial distance "r", then its linear velocity resulting from pure rotation is given by : ⇒ v = ωr Substituting value of ω from earlier equation, we can obtain the velocity of a particle inside the rotating disk as : v = (vC r)/R (2) where "r" is the linear distance of the position occupied by the particle from the axis of rotation.

Pure translation For pure translation, we consider that the rotating disk is not rotating at all. Each particle of the disk is translating with linear velocity that of center of mass, " vC". Unlike the case of pure rotation, each of the particle - whether situating on the rim or within the disk - is moving with same velocity. In the figure, we have shown the linear velocities of particles with appropriate vectors, occupying four positions on the rim.

7.13.2Kinetic energy of rolling disk We can determine kinetic energy of the rolling disk, considering it to be the combination of pure rotation and pure translation. Mathematically, K=KR+KT K = 1/2 ICω2 + 1/2 MvC2

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