Class 11 Exam > Class 11 Videos > Numericals: Center of mass - System of Particles & Rotational Motion
Numericals: Center of mass - System of Particles & Rotational Motion Video Lecture - Class 11
FAQs on Numericals: Center of mass - System of Particles & Rotational Motion Video Lecture - Class 11
| 1. What is the concept of the center of mass in a system of particles? |  |
Ans. The center of mass is a point in a system of particles that represents the average position of all the particles in the system. It can be thought of as the balance point or the point at which the system would remain balanced even if external forces were applied.
| 2. How is the center of mass of a system of particles calculated? | |
Ans. The center of mass of a system of particles can be calculated by taking the weighted average of the positions of all the particles, where the weight is given by the mass of each particle. Mathematically, the center of mass (x_cm, y_cm, z_cm) can be calculated using the formula:
x_cm = (m1x1 + m2x2 + m3x3 + ... + mnxn) / (m1 + m2 + m3 + ... + mn)
y_cm = (m1y1 + m2y2 + m3y3 + ... + mny) / (m1 + m2 + m3 + ... + mn)
z_cm = (m1z1 + m2z2 + m3z3 + ... + mnz) / (m1 + m2 + m3 + ... + mn)
where m1, m2, m3, ..., mn are the masses of the particles, and x1, x2, x3, ..., xn, y1, y2, y3, ..., yn, z1, z2, z3, ..., zn are the coordinates of the particles.
| 3. What are some applications of the concept of center of mass in real life? | |
Ans. The concept of center of mass has various applications in real life. Some examples include:
- Balancing objects: Understanding the center of mass helps in balancing objects, such as in the construction of stable furniture or structures.
- Sports: Athletes often utilize the concept of center of mass to maintain balance and stability during movements, such as in gymnastics or figure skating.
- Vehicle design: Engineers consider the center of mass while designing vehicles to ensure stability and prevent tipping over.
- Astronomy: The center of mass helps in studying the motion and interactions of celestial bodies, such as planets, moons, and galaxies.
- Robotics: Robots are designed with a stable center of mass to ensure balance and prevent falling during movements.
| 4. Is the center of mass always located within an object? | |
Ans. No, the center of mass is not always located within an object. In some cases, the center of mass may lie outside the physical boundaries of an object. This can happen when the distribution of mass within the object is uneven or when there are external forces acting on the object. However, regardless of its position, the center of mass still represents the average position of the particles in the system.
| 5. How does the concept of center of mass relate to rotational motion? | |
Ans. The concept of center of mass is closely related to rotational motion. When an object undergoes rotational motion, it rotates around an axis passing through its center of mass. This is because the center of mass represents the point where the mass of the object can be considered to be concentrated. The rotational motion of an object can be analyzed using concepts such as moment of inertia and angular momentum, which are calculated with respect to the center of mass. Understanding the center of mass is crucial in studying and predicting the rotational behavior of objects.
Text Transcript from Video
hello friends this video on system of
particles and rotational motion part six
is brought to you by example.com move
move from example now keeping all these
points in mind let us try to calculate
the location of centre of mass for a few
objects so let us look at this first
problem 1 the problem says give the
location of the center of mass of a
sphere cylinder ring and Hume each of
uniform mass density so uniform mass
density plays the key role here what do
you mean by doing for mass density that
means the mass is spread uniformly
throughout the object that means we are
talking about homogeneous bodies now we
know that for homogeneous bodies the
center of mass will lie at their
geometric center so that means for this
fear this will be the center of mass for
this cylinder again the geometrical
center of the cylinder similarly for
this thing the geometrical center is
this one right similarly for a cube even
if you consider a cube like this the
center of mass will be concentrated at
this point so here we are talking about
homogeneous objects so therefore center
of mass will lie at their geometric
centers so the first part is done
the next part it asks does the center of
mass of a body necessarily lie inside
the body so what that you can answer
looking at these figures do you think
that the center of mass will always lie
necessarily inside the body no that's
because if you look at this room this is
the center of mass but the body of this
ring is only this part right I mean you
can pick out your bangle and see the
example the sending of mass does not lie
inside the bangle right because the
bangle is it is hollow come inside it is
just the boundary so if I say that
something lies into the black mass lies
inside the ring that means the center of
mass mass should lie somewhere here
inside the ring so the center of mass
lies outside the ring right so it is not
necessary that the center of mass will
always lie inside
the body for example if you look at this
fear if the sphere is a solid sphere
then we will say that yes the center of
mass lies inside the body of respondents
fear but if we are talking about the
whole new sphere which is empty from
inside in that case we will see that the
center of mass does not length inside
the corner of the scale Hey are you
getting my point so the ring was the
best example which I mentioned but
otherwise in case of the sphere or the
cylinder here the center of mass is
lying inside the body let us look at the
next problem it says that in HCl
molecule the separation between the
nuclei of the two atoms is one point two
seven and strong so what are the two
atoms in case of HCl hydrogen atom and
chlorine atom so it sees that the
separation between the nuclei of these
two atoms is one point two seven and
structure that means this distances one
point two seven find the approximate
location of the center of mass of the
molecule given that a chlorine atom is
thirty five point five times as massive
as hydrogen atom and nearly all the mass
of an atom is concentrated in its
nucleus let's pretty fine lady because
in case of an atom the nucleus is the
one which has the maximum mass because
the electrons are very small mass so now
let us suppose let us suppose that this
is my center of mass so let C denote my
center of mass so this is point C so
what is the distance of fluorine from C
let us suppose the distance of chlorine
from the center of mass is X and the
floating of hydrogen L would be one
point two seven minus X because this
whole distance is one point two seven
right and according to the definition of
center of mass so how do we define the
center of mass the center of mass will
be defined by its own coordinate X right
because here we do not have any
involvement of Y axis or z axis so this
H will be given as m1 x1
M 2 X 2 divided by M 1 plus M 2 that is
how we determine the center of mass so
what is my let us suppose 1 denotes the
mass of hydrogen and 2 denotes the mass
of chlorine so M 1 is the mass of
hydrogen let us suppose that mass of
hydrogen is M let us suppose this is M
so what would be the mass of floating
input to 35.5 times of M so this will be
M into X 1 what is X 1 1 point 2 7 minus
X plus thirty five point five M into X 2
what is X 2 that is X this divided by M
1 plus M 2 that is M plus thirty five
point five M so what would be this so
this is your center of mass right we
have assumed that this is our center of
mass and we have also assumed that this
is the origin by considering this was
originally we have considered this as X
and this s one point two seven minus X
so if we have considered this as origin
then this should be equal to zero right
so now if you solve this you get one
point two seven minus X is equal to
minus X 5.5 X so from this we get X is
20 minus zero point zero seven and
strong so this is the distance so this
is the value of X so this will be my X
so therefore we can see that the center
of mass lies at a distance of zero point
zero three seven and strong from
chlorine atom so this is the position of
the center of mass so what did we do
here we assume that let us suppose that
the center of mass is located at this
position and we also assume that this is
the origin if we consider this at the
origin then we can measure the distances
of these two atoms from the origin at
the same time my center of mass
coordinate to be equal to 0 so we
equated it to zero and that is how we
calculate
the value of x and found out the
approximate location of the center of
mass let us now look at the third
problem it says that find the center of
mass of a uniform l-shaped lamina with
dimensions as shown the mass of the
lambda is 3 kgs so if you look at this
lamina this is the l-shaped lamina so
this l-shaped lamina actually consists
of 3 squares 1 2 & 3 so these are the
three squares which together form the
lamina and you can see that the center
of mass of each square lines at their
geometric centers because it is a
uniform lamina rattles the mass
distribution is uniform now for
homogeneous bodies so for these squares
this P 1 C 2 C 3 are their centers
therefore their center of mass and
located at C 1 C 2 and C 3 right so let
us first find out the coordinates of the
center of masses by looking at this
figure we can say that what is the
coordinate of C 1 it will be on x-axis
it is 1 by 2 because this entire 2 is 2
that means this is 1 so this will be 1
by 2 so every 1 by 2 and what is y axis
or Y axis this is 2 definitely this is 1
so that means this is again 1 by 2 C 2
for C 2 this is will be one solution
this will be 1 plus 1 by 2 that means
the x coordinate is 3 by 2 what about
the y coordinate it is 1 by 2 for C 3 /
c 3 the x coordinate is again 1 by 2 but
the y coordinate is 1 plus 1 by 2 that
is 3 by 2 and what about the masses of
each of these C 1 C 2 C 3 M 1 M 2 and M
3 each will be equal to 1 kg because the
total mass is 3 pages that means 1 kg
plus 1 kg
one cage right so each of the masses is
three caging so now we have to calculate
the center of mass so let us assume that
the center of mass is given by the
coordinates x and y so here we have got
X 1 y 1 this is X 2 y 2 and this is X 3
y 3 right so therefore X as per
definition is nothing but summation of
m-i X I divided by the total mass so
this will be equal to M 1 X 1 plus M 2 X
2 plus M 3 X 3 divided by the total mass
so what is the total mass total mass of
three pages so this will be equal to L 1
is 1 in P 1 by 2 plus L 2 is again 1
into X 2 is 3 by 2 plus M 3 is 1 into X
3 that is 1 by 2 divided by 3 so this
comes out as 5 by 6 meters so this is
the value of the X 1 so similarly we
differentiate the value of the y
coordinate which will come out as M 1 y
1 plus M 2 by 2 plus M 3 rightly divided
by the total mass capital 3 right so
this Y will come out to be again 5 by 6
meters so therefore we can see that the
center of mass of the lamina is given by
the coordinates 5 by 6 meters and 5 by 6
meters so this is how we determined the
center of mass of a rigid body
next we'll understand the concept of
center of mass and how do we calculate
the center of mass of a system of
particles how do we calculate the center
of mass of a rigid body now with this
thank you please visit exam feel calm to
watch free educational videos try 3
online tests get the best quality study
materials study from the best tutors and
mentor
and much more thank you once again
particles and rotational motion part six
is brought to you by example.com move
move from example now keeping all these
points in mind let us try to calculate
the location of centre of mass for a few
objects so let us look at this first
problem 1 the problem says give the
location of the center of mass of a
sphere cylinder ring and Hume each of
uniform mass density so uniform mass
density plays the key role here what do
you mean by doing for mass density that
means the mass is spread uniformly
throughout the object that means we are
talking about homogeneous bodies now we
know that for homogeneous bodies the
center of mass will lie at their
geometric center so that means for this
fear this will be the center of mass for
this cylinder again the geometrical
center of the cylinder similarly for
this thing the geometrical center is
this one right similarly for a cube even
if you consider a cube like this the
center of mass will be concentrated at
this point so here we are talking about
homogeneous objects so therefore center
of mass will lie at their geometric
centers so the first part is done
the next part it asks does the center of
mass of a body necessarily lie inside
the body so what that you can answer
looking at these figures do you think
that the center of mass will always lie
necessarily inside the body no that's
because if you look at this room this is
the center of mass but the body of this
ring is only this part right I mean you
can pick out your bangle and see the
example the sending of mass does not lie
inside the bangle right because the
bangle is it is hollow come inside it is
just the boundary so if I say that
something lies into the black mass lies
inside the ring that means the center of
mass mass should lie somewhere here
inside the ring so the center of mass
lies outside the ring right so it is not
necessary that the center of mass will
always lie inside
the body for example if you look at this
fear if the sphere is a solid sphere
then we will say that yes the center of
mass lies inside the body of respondents
fear but if we are talking about the
whole new sphere which is empty from
inside in that case we will see that the
center of mass does not length inside
the corner of the scale Hey are you
getting my point so the ring was the
best example which I mentioned but
otherwise in case of the sphere or the
cylinder here the center of mass is
lying inside the body let us look at the
next problem it says that in HCl
molecule the separation between the
nuclei of the two atoms is one point two
seven and strong so what are the two
atoms in case of HCl hydrogen atom and
chlorine atom so it sees that the
separation between the nuclei of these
two atoms is one point two seven and
structure that means this distances one
point two seven find the approximate
location of the center of mass of the
molecule given that a chlorine atom is
thirty five point five times as massive
as hydrogen atom and nearly all the mass
of an atom is concentrated in its
nucleus let's pretty fine lady because
in case of an atom the nucleus is the
one which has the maximum mass because
the electrons are very small mass so now
let us suppose let us suppose that this
is my center of mass so let C denote my
center of mass so this is point C so
what is the distance of fluorine from C
let us suppose the distance of chlorine
from the center of mass is X and the
floating of hydrogen L would be one
point two seven minus X because this
whole distance is one point two seven
right and according to the definition of
center of mass so how do we define the
center of mass the center of mass will
be defined by its own coordinate X right
because here we do not have any
involvement of Y axis or z axis so this
H will be given as m1 x1
M 2 X 2 divided by M 1 plus M 2 that is
how we determine the center of mass so
what is my let us suppose 1 denotes the
mass of hydrogen and 2 denotes the mass
of chlorine so M 1 is the mass of
hydrogen let us suppose that mass of
hydrogen is M let us suppose this is M
so what would be the mass of floating
input to 35.5 times of M so this will be
M into X 1 what is X 1 1 point 2 7 minus
X plus thirty five point five M into X 2
what is X 2 that is X this divided by M
1 plus M 2 that is M plus thirty five
point five M so what would be this so
this is your center of mass right we
have assumed that this is our center of
mass and we have also assumed that this
is the origin by considering this was
originally we have considered this as X
and this s one point two seven minus X
so if we have considered this as origin
then this should be equal to zero right
so now if you solve this you get one
point two seven minus X is equal to
minus X 5.5 X so from this we get X is
20 minus zero point zero seven and
strong so this is the distance so this
is the value of X so this will be my X
so therefore we can see that the center
of mass lies at a distance of zero point
zero three seven and strong from
chlorine atom so this is the position of
the center of mass so what did we do
here we assume that let us suppose that
the center of mass is located at this
position and we also assume that this is
the origin if we consider this at the
origin then we can measure the distances
of these two atoms from the origin at
the same time my center of mass
coordinate to be equal to 0 so we
equated it to zero and that is how we
calculate
the value of x and found out the
approximate location of the center of
mass let us now look at the third
problem it says that find the center of
mass of a uniform l-shaped lamina with
dimensions as shown the mass of the
lambda is 3 kgs so if you look at this
lamina this is the l-shaped lamina so
this l-shaped lamina actually consists
of 3 squares 1 2 & 3 so these are the
three squares which together form the
lamina and you can see that the center
of mass of each square lines at their
geometric centers because it is a
uniform lamina rattles the mass
distribution is uniform now for
homogeneous bodies so for these squares
this P 1 C 2 C 3 are their centers
therefore their center of mass and
located at C 1 C 2 and C 3 right so let
us first find out the coordinates of the
center of masses by looking at this
figure we can say that what is the
coordinate of C 1 it will be on x-axis
it is 1 by 2 because this entire 2 is 2
that means this is 1 so this will be 1
by 2 so every 1 by 2 and what is y axis
or Y axis this is 2 definitely this is 1
so that means this is again 1 by 2 C 2
for C 2 this is will be one solution
this will be 1 plus 1 by 2 that means
the x coordinate is 3 by 2 what about
the y coordinate it is 1 by 2 for C 3 /
c 3 the x coordinate is again 1 by 2 but
the y coordinate is 1 plus 1 by 2 that
is 3 by 2 and what about the masses of
each of these C 1 C 2 C 3 M 1 M 2 and M
3 each will be equal to 1 kg because the
total mass is 3 pages that means 1 kg
plus 1 kg
one cage right so each of the masses is
three caging so now we have to calculate
the center of mass so let us assume that
the center of mass is given by the
coordinates x and y so here we have got
X 1 y 1 this is X 2 y 2 and this is X 3
y 3 right so therefore X as per
definition is nothing but summation of
m-i X I divided by the total mass so
this will be equal to M 1 X 1 plus M 2 X
2 plus M 3 X 3 divided by the total mass
so what is the total mass total mass of
three pages so this will be equal to L 1
is 1 in P 1 by 2 plus L 2 is again 1
into X 2 is 3 by 2 plus M 3 is 1 into X
3 that is 1 by 2 divided by 3 so this
comes out as 5 by 6 meters so this is
the value of the X 1 so similarly we
differentiate the value of the y
coordinate which will come out as M 1 y
1 plus M 2 by 2 plus M 3 rightly divided
by the total mass capital 3 right so
this Y will come out to be again 5 by 6
meters so therefore we can see that the
center of mass of the lamina is given by
the coordinates 5 by 6 meters and 5 by 6
meters so this is how we determined the
center of mass of a rigid body
next we'll understand the concept of
center of mass and how do we calculate
the center of mass of a system of
particles how do we calculate the center
of mass of a rigid body now with this
thank you please visit exam feel calm to
watch free educational videos try 3
online tests get the best quality study
materials study from the best tutors and
mentor
and much more thank you once again
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