Physics-PHY101-Lecture 12

PHYSICS OF MANY PARTICLES

Introduction to center of mass:

Everybody is made up of particles and every particle has a direction with every other particle, that
it either pulls it towards itself or pushes it. There are some quantities about which if we have the
information then we can say something about the system as a whole which means they include the
center of mass. So where the center of the mass will be, there will be a kind of concentration of
mass.

Let's say there is an elephant. An ant moves on top of the elephant. It will not make much difference
if the ant moves here or there. The mass of the elephant is so much that the center of mass will be
the center of elephant. We would like to define the center of mass as,

A system's or object's center of mass is the location where the mass is considered to be
concentrated. Simply, it's the average position of the mass distribution.

Mathematically for two masses the center of mass rcm is:

+ m 2r 2
r = m1r 1
cm
m +m
1 2

Figure 12.1. Distribution of masses ‘m1 and m2’ with their position vectors ‘r1 and r2’.
The center of mass equation can be written in two or three dimensions, depending on the given
situation in question. In the case of two dimensions, its coordinates will be x and y; in the case of
three dimensions, there will be x, y, and z. Equation of center of mass for two dimension is:

+ m2 x 2
x = m1 x 1
m +m
cm
1 2

=
m y +m y
1 2
y 1 2

m +m
cm
1 2

The center of mass will lie between two masses if their masses are equal (m1 = m2 = m), and closer
to the heavier mass if one mass is significantly greater than the other. If M is heavier than m, the
center of mass is near mass M as can be seen from figure 12.2.

Figure 12.2. Center of mass alignment for unequal masses. Equal masses (m1 = m2 = m) result in a
center of mass equidistant between them. For unequal masses (M > m), the center of mass is closer to
the heavier mass.

Let’s take another example. Consider two bodies of equal masses, one is placed at x = 2 m and
other is placed at x = 6 m as shown in figure 12.3, then center of mass will be ,
 Figure 12.3. Two bodies of equal masses are positioned at x = 2 m and x = 6 m. The
center of mass location is determined by their equilibrium.

mx1 + mx2
xcm =
m+m

m(2 meter ) + m(6 meters )

x cm
=
2m
8m
x cm = 2m = 4m (4 meters)
Now consider two bodies of unequal masses. One is a lighter ‘mass m’ placed at x = 6 m and other
is a heavier ‘mass 3m’ placed at x = 2 m as shown in figure 12.4.

Figure 12.4. A lighter mass 'm' is positioned at x = 6 m and a heavier mass '3m' at x = 2
m. The center of mass is influenced by the masses and their respective positions.

(3m) x1 + mx2 (3m)2 + 6m 12m

xcm = = = = 3 meters
3m + m 4m 4m

For N masses, the center of mass is:

 m1r1 + m2 r2 + + mN rN
rcm =
m 1 + m2+ + + mN

rcm =
1
M
( m r )n n

Figure 12.5: There is a point CM which we call center of mass, from the origin O, vector r be the
position vector.

For symmetrical object, the center of mass is easy to guess as can be seen from the figures below.

(a) (b) (c)

In figure (a), the center of mass is in the center of star and in (b) the center of mass is in the center
of cylinder if the material of cylinder will be uniform and in figure (c) if it’s a uniform sphere then
its center of mass is also in the center of sphere.

In order to understand that why the center of mass is useful concept, we have to apply newton’s
law. The velocity and acceleration of the center of mass can be calculated by the definition of the
center of mass and differentiating it w.r.t time.

vcm =
drcm
dt
=
1
M
( m v )n n  rcm =
1
M
( m r )
n n
 acm =
dvcm
dt
=
1
M
( m a )
n n

It is known as the acceleration of a single point, and we know that a particle possesses acceleration
only when a force act on it. Here, the force can act for two reasons.

• One reason is that the particle is in external field. For example, the particle is inside the
gravity and the gravity is pulling it down or up or in some direction.
• The second reason is that the other particles exert force and push or pull the particle in
different directions.

The two types of forces that act on every particle are divided into external and internal forces.
Therefore, the acceleration of each particle is due to the internal and external forces, but the internal
force gets cancelled (because action and reactions are equal and opposite) and only external force
remains.

(
Macm =  Fn =  Fext + Fint )
F ext = Macm

In conclusion, the center of mass moves in such a way to concentrate the whole mass on it.

A center of mass is a point but it’s not necessarily for a mass to be actually there. It's a common
misconception that there's actually something. It is only appearing that all the masses concentrated
there but when Newton's Law is applied, this imaginary point moves just like a mass that is
concentrated at one point.

The center of mass for a combination of objects is the average center of mass location of objects.
The center of mass can be outside the body; it does not have to be inside for all the structures as
can be seen from figure 12.6.
 Figure 12.6. The center of mass represents the average location, and it can be located
outside the physical boundaries of the objects.

Rotational Energy of Rigid Bodies:

Where there is movement there will be energy called kinetic energy. Motion is not only in one
direction it can also be rotational motion. Consider a rigid body rotating about a fixed axis as
shown in figure 12.7.

Figure 12.7. Rigid body rotation about a fixed axis.

Now let’s know what a rigid body is. A body which has no elasticity, in which all its particles
move together, such as wheel. If you calculate the total kinetic energy of a rigid body, then it is
the sum of all the kinetic energies of each body.

Total kinetic energy is:

 1 1 1
K= m1v12 + m2 v2 2 + m3v32 +
2 2 2
v = r

For a rigid body or an inelastic body, all the particles are moving together, which means that each
particle has the same angular velocity ‘ω’ as the other particles, when we add all the K.E we get,

1 1 1
K= m1r12 2 + m2 r2 2 2 + m3 r32 2 +
2 2 2

K=
1
2
(  mi ri 2 )  2

Where  mi ri 2 is called as ‘moment of inertia I’.

Rotational inertia:

K=
1
2
(  mi ri 2 )  2

⸫ I =  mi ri 2
1 2
K = I
2

This implies that we will take the mass of each body and use the square of the distance to calculate
each body's moment of inertia. Then, we will add up all of the body's moments of inertia to get the
total moment of inertia. When a body moves on straight line than its KE is:

1
K= Mv 2 ⸫ v is the linear velocity.
2

Problem:
Two particles m1 and m2 are connected by a light rigid rod of length L. neglect the mass of rod,
find the rotational inertia I of this system about an axis perpendicular to the rod and at a distance
x from m1.

Solution:

I = I1 + I 2
I = m1 x 2 + m2 ( L − x )
2

For what x, is I the largest?

dI d
=
dx dx
(
m1 x 2 + m2 ( L − x )
2
)
dI
= m1 ( 2 x ) + m2 2 ( L − x )( −1)
dx
dI
For a maxima, =0
dx

0 = 2m1 x − 2m2 ( L − x )
0 = 2m1 x − 2m2 L + 2m2 x
( 2m1 + 2m2 ) x = 2m2 L
2m2 L m2 L
x= =
2(m1 + m2 ) (m1 + m2 )

Conclusion: At this distance, the system of particle possess the highest value of moment of
inertia. If m1 = m2 then x = L/2 for maxima.

Problem:
Three particles of masses m1 (2.3 kg), m2 (3.2 kg) and m3 (1.5 kg) are at the vertices of a
triangle.

Part-I: Find the rotational inertia about axes

perpendicular to the xy plane and passing
through each of the particles.

Part-II: What is the moment of inertia about

the center of mass?

Solution:

Moment of inertia about each axis of rotation =?

When passing through particle 1:
I = I1 + I 2 + I 3 = mr12 + mr22 + mr32
I = 2.3kg (0) + 3.2kg (3) 2 + 1.5kg (4) 2 = 52.8kg
When passing through particle 2:
I = 2.3kg (3) 2 + 3.2kg (0) + 1.5kg (5) 2 = 58.2kg
When passing through particle 3:
I = 2.3kg (4) 2 + 3.2kg (5) 2 + 1.5kg (0) = 116.8kg

Part-II: What is the moment of inertia about the

center of mass?
Lets find the center of mass first.
Since, masses are in xy-plane so,
rcm = xcm i + ycm j
m1 x1 + m2 x2 + m3 x3 m1 (0) + m2 (0) + m3 (4) 6
xcm = = = = 0.86m
m1 + m2 + m3 2.3 + 3.2 + 1.5 7
m1 y1 + m2 y2 + m3 y3 m1 (0) + m2 (3) + m3 (0) 9.6
ycm = = = = 1.37m
m1 + m2 + m3 2.3 + 3.2 + 1.5 7

Using pythagoras theorem,

= ( 0.86 ) + (1.37 ) = 2.62m 2
2 2
r12 = xcm
2
+ ycm
2

+ ( y2 − ycm ) = ( 0.86 ) + ( 3 − 1.37 ) = 3.40 m 2

2 2 2
r22 = xcm
2

r32 = ( x3 − xcm ) + ycm = ( 4 − 0.857 ) + (1.37 ) = 11.74m 2

2 2 2 2

I cm =  mi ri 2 = m1r12 + m2 r22 + m3r32

I cm = 2.3* 2.62 + 3.2*3.40 + 1.5*11.74
I cm = 34.5 kg m 2

This problem is for three masses, you may solve it for infinite masses. We use integration if we
have infinite masses. If we divide a body into small parts, if mass of a small parts is dm. Take the
distance r from the center and then take square of distance and then multiply with dm and then
sum up over all the parts of the body that make up. This is called integration, and the total moment
of inertia is the integration or the integral of the entire body of r square times dm.

For solid bodies:

I =  r 2 dm

Hoop about cylinder axes:

I =  r 2 dm r = R= Fix
I = R 2  dm
I = MR 2
 Figure 12.8. Hoop rotation about cylinder axis.

Solid plate about cylinder axes:

I =  r 2 dm
M M
  = = 2
Area  r
dm dm
if   = =
dA 2 rdr
dm = 2 rdr 
R
I =  r 2 .2 rdr 
0
R R
I =  2 r dr   = 2   r 3dr
3

0 0

4 R
r
I = 2 
4 0

I=
1
2
(  R 4   ) = ( R 4 ) 2
1
2
M
R
1
I = MR 2
2

Solid sphere about diameter:

2
I= MR 2
5
 Figure 12.9. Solid sphere rotation about diameter.

For a hollow sphere, the mass is concentrated towards the outside, and you would get the result
that a hollow sphere of mass m will have a greater moment of inertia than a sphere of mass m.

The moment of inertia measures how much energy is generated inside an object when you spin it.
For example as shown in figure 12.10 (a) and (b).

(a) Solid cylinder or disk about cylinder axis

1
I = MR 2
2

(b) Solid cylinder or disk about central diameter:

1 1
I = MR 2 + ML2
4 12

Figure 12.10. (a) solid cylinder or disk about cylinder axis, (b) solid cylinder or disk
about central diameter .

Rectangular plate about central axis:

 I=
1
12
(
M a 2 + b2 )

Figure 12.11. Rectangular plate about central axis.

So in conclusion, if an object rotates, its kinetic energy becomes equal to the ½ I ω2, but the
question arises here is that, why the object rotates? Of course, force helps to rotate the object, but
it depends on where the force is applied. For example, we have wrenches of various sizes. Because
of torque, when we use a small-length wrench, it is difficult to tighten the nut; however, when we
use a large-length wrench, the nut is easily tightened. A wrench with a long length has higher
torque than one with a little length.

Now we define torque mathematically,

 = r F

Where ‘r’ is the distance at which force acts. If we take its magnitude, we have

 = rF sin 

And θ is the angle between r and F as shown in figure 12.12.

 Figure 12.12. Illustration of torque, where 'r' is the distance at which force (F) acts, and θ
is the angle between r and F.

Work can be done wherever a force occurs and when it acts and moves a distance S, the force into
a distance is equal to work as shown in figure 12.13. So F.ds is the amount of work done.

dW = F .ds

Figure 12.13. Work (W) is the product of force (F) and displacement (ds), where the
force acting over a distance results in the performance of work.

So suppose there are several bodies and forces acting on them. Now the work done is,
 dW = F .ds = F cos  ds
= ( F cos  )( rd )
=  d

For all particles,

dWnet = ( F1 cos 1 ) r1d + ( F2 cos  2 ) r2 d + + ( Fn cos  n ) rn d

= ( 1 +  2 + +  n ) d

See, there's only one angle here, and that's because all the particles move together because it's a
rigid body.

Wnet = (  ext ) d = (  ext ) dt

When we differentiate the K.E with respect to  ,

1 
K =  I 2 
2 
dK d 1 2 1
=  I   = I (2 ) = I 
d d  2  2
d
dK = I  d  =  d =  dt
dt
dK = ( I )  dt

Remember that α is the angular acceleration.

dWnet = dK

Work has been done by you in rotating it so that work is converted into kinetic energy. Then the
result is the total external torque and is equal to moment of inertia multiplied by angular
acceleration,

 ext = I

This equation is analogous to the Newton’s law F = ma.

Before continuing we should know where translational and rotational motions are similar.
‘x’ is displacement at the one side and Ø is angular displacement on the other side. As x increases
with the time and we take the derivative dx/dt, likewise Ø increases with time and we take the
derivative from there we get the angular speed.

Translational motion Angular motion

x, M , I

dx d
v= =
dt dt

dv d
a= =
dt dt

In term of dynamics, the moment of inertia has exactly the same role for rotational motion as that
of mass in translational motion.

Translational Rotational motion

F = Ma  = I

W =  Fdx W =   d

1 1 2
K= Mv 2 K= I
2 2

Combined Rotational and Translational Motion:

In order to understand this type of motion, let’s take the example of a car. Take the wheel of a car,
put a mark on the wheel, and put a mark on its rim. The car goes in a straight line, but at the same
time, its wheel rotates. So, this is an example of combined rotational and translational motion.

Now we study it in detail, there is a body with two vectors on it. One vector is of center of mass
and the second is at point p, point P is any random point on a body and a vector which goes from
the center of mass to P is called ri ' as shown in figure 12.14.
 Figure 12.14. Body with center of mass vector and position vector at point P. The vector
from the center of mass to a random point p on the body is defined as the position vector
𝒓′𝒊 .

If we take out the total kinetic energy of it, then it will have two parts, one part is the kinetic energy
of the center of the mass and the other part is due to its rotation.

1 1
K= Mvcm 2 + I cm 2
2 2

Remember, we didn't take a regular shape here. The body is rolling along its path in addition to
traveling in a straight line. We now aim to find out if it is indeed the case that this body possesses
two different kinds of kinetic energy. Let's prove it.

First of all, we will extract the kinetic energies of all its different particles, then we divide the
velocity of every particle into two parts. One is the part of center of mass and other is due to
rotation.
 1 1
K =  mi vi2 =  mi vi .vi
2 2
1
 2 mi ( vcm + vi ) . ( vcm + v)
 2 mi ( v 2cm + 2vcm .vi + vi 2)
1

 mi vcm .vi = vcm . mi vi

 p =  m v = Mv
i i i cm

 =0
vcm

 = 0 in the center of mass frame.

vcm

1 1
K =  mi v 2 cm +  mi v2
2 2
1 1
= Mv 2 cm +  mi ri2 2
2 2
1 1
K = Mv 2 cm + I cm 2
2 2

Here, we emphasized that it is a rigid body that moves along. However, we cannot apply this to a
non-rigid body in such a way that, for example, rotating a bucket of water will cause the water
inside to rotate at one speed while the water attached to the bucket rotates at a different speed,
indicating that the bucket is not a rigid body.

Rolling without slipping:

Take a rigid body with a center of the mass speed (vcm) moving toward the right side and this is
just translational motion. And on the other hand it is just rotational motion that is rotating along
the fix center and its speed is Rω on their edges as shown in figure 12.15.
 Translational motion Rotational motion

Figure 12.15. The body exhibits translational motion with center of mass speed (vcm) to
the right and rotational motion about a fixed center with speed Rω.

It is possible that it is rolling and rotating together, so it is a combination of translational plus

rotational motion. Here point B is attached to the ground so it is at rest. It is moving at the speed
of Vcm, and the top point is moving at two times the velocity of center of mass as shown in figure
12.16.

Translational + Rotational motion

Figure 12.16. Rigid body with combined translational (vcm) and

rotational (Rω) motion about a fixed axis.

If we find the total energy then we see the result of rolling without slipping is,
 vcm = R
1 1  v 2 cm 
K = Mv cm + I cm  2 
2

2 2  R 
1 1
K= MR 2 2 + I cm 2
2 2

We didn’t say that this is a sphere or a hoop or the other shape we only discuss it as a rigid body.

Example:

In the absence of friction, energy is conserved, meaning that the same amount of energy at the
beginning remains at the end and since we know that energy is divided into two parts; kinetics
energy of center of mass plus rotational motion so from this we can solve many types of questions.

Now we take the example of a body rolling down a slope as shown in figure 12.17. For example a
hoop which has the moment of inertia ½ mR2. Now we want to know how fast it will be when it
leaves the slope.

Figure 12.17. Rolling Motion of a Hoop on a Slope.

Solution:

Here we use the conversation of energy, in the start it has potential energy but later its loss the
potential energy and only kinetic energy remains. When body leaves the slop its K.E is,
 At starting point, object possess gravitational P.E = Mgh
As it moves down, it possess rotational and translational K.E, which is:
1 1
K = Mv 2 cm + I cm 2
2 2
2
1 11  v 
Mgh = Mvcm
2
+  MR 2  cm   vcm = r
2 22  R 
2
vcm v2 3v 2
gh = + cm = cm
2 4 4
4
vcm = gh
3

Summary: Kinetic energy consists of two parts, one related to the center of mass and one related
to its rotation. Center of mass and moment of inertia are two properties associated with a body and
which define the body. What we've discussed is very important for mechanical engineers, people
who design machines that have rotating objects inside, whether they're aircraft engineers or
automobile engineers or others. We will go far with these concepts, which are essential in every
other subject that we have learnt.