Class 12 Exam > Class 12 Videos > L25 : Circular motionroblems - Laws of motion, Physics, Class 11
L25 : Circular motionroblems - Laws of motion, Physics, Class 11 Video Lecture - Class 12
FAQs on L25 : Circular motionroblems - Laws of motion, Physics, Class 11 Video Lecture - Class 12
| 1. What are the laws of motion? |  |
Ans. The laws of motion are three fundamental principles stated by Sir Isaac Newton that describe the relationship between the motion of an object and the forces acting upon it. These laws are:
1. Newton's First Law of Motion (Law of Inertia): An object at rest will stay at rest, and an object in motion will stay in motion with the same velocity unless acted upon by an external force.
2. Newton's Second Law of Motion: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. It can be mathematically expressed as F = ma, where F is the net force, m is the mass of the object, and a is the acceleration.
3. Newton's Third Law of Motion: For every action, there is an equal and opposite reaction. This law states that if an object A exerts a force on object B, then object B exerts an equal and opposite force on object A.
| 2. What is circular motion? | |
Ans. Circular motion refers to the movement of an object along a circular path. In circular motion, the object continuously changes its direction while maintaining a constant distance from a fixed point called the center of the circle. This type of motion involves both linear motion (tangential to the circle) and radial motion (directed towards or away from the center of the circle). Examples of circular motion include a car traveling around a roundabout or a planet orbiting around the sun.
| 3. How is circular motion related to the laws of motion? | |
Ans. Circular motion is related to the laws of motion through the application of the second law of motion. According to Newton's second law, the net force acting on an object is directly proportional to its acceleration. In circular motion, there is a centripetal force acting towards the center of the circle, which keeps the object moving in its circular path. This centripetal force is responsible for the object's acceleration towards the center of the circle. Thus, the second law of motion provides the mathematical relationship between the centripetal force, mass of the object, and its acceleration in circular motion.
| 4. What is the difference between uniform circular motion and non-uniform circular motion? | |
Ans. In uniform circular motion, the speed of the object remains constant throughout its circular path, while in non-uniform circular motion, the speed varies. In uniform circular motion, the object experiences a constant centripetal force that keeps it moving in a circle at a constant speed. On the other hand, in non-uniform circular motion, the object may experience varying centripetal forces, resulting in changes in speed along its circular path.
| 5. Give an example of circular motion in everyday life. | |
Ans. An example of circular motion in everyday life is a washing machine. When a washing machine is in operation, the clothes inside undergo circular motion. The drum of the washing machine rotates in a circular path, causing the clothes to move along the walls of the drum. This circular motion helps in the effective cleaning of the clothes by agitating them with detergent and water.
Text Transcript from Video
hello friends this video on laws of
motion part 25 is brought to you by exam
feel calm no more fear from exam please
make sure that you have watched all the
videos from part 1 part 24
before going ahead with power 25 start
with solving some new mannequins the
first problem says find the angle
through which a cyclist vents from the
vertical when he covers a circular path
of thirty four point three meter
self-confidence in route over 22 seconds
that means the self a cyclist is trying
to cover a circular part whose
circumference is given as thirty four
point three meters that means what is
the circumference of a circle it is 2 pi
add so value of 2 PI R is given as 34
point three meters and the time taken to
cover this much of distance is T is
equal to root over 22 seconds so
basically this 2 PI R is nothing but the
distance covered by the cyclist so but
through which angle he needs to bend
what is the angle that he makes with a
vertical so let us try to draw the
diagram so that he can able to visualize
the situation let us suppose this is my
vertical because it says that the
cyclist bends from the vertical let us
suppose this is the cyclist is just a
rough figure let us say this is the
cyclist like so he is making let us
suppose that he is making some angle
theta so we need to find the value of
this theta let us suppose he is making
some angle theta and he is trying to
take a curve now what will be the
different forces acting on the cyclist
because he is taking a turn somewhat
like this it's somewhat like this so
mass of the body will be in the I mean
weight of the body with in the downward
direction that is mg if M is the mass of
the cyclist now what what will be the
direction of the normal reaction normal
reaction is always by
when dick Euler to the surface of
contact now what is the surface of
contact head this is the surface of
contact if you see the cyclist is in
contact with the road at this point so
this is my surface of contact therefore
the normal reaction would be
perpendicular to this so this would be
the direction of the normal reaction
suppose if this is Theta so along this
direction what would be the component of
normal reaction this would be n cos
theta right so this would be n cos theta
therefore this side it would be n sine
theta correct now what is forcing the
cyclist to move in a circular path
obviously centripetal force centripetal
force would be along this direction
because it always act towards center
that means the centripetal force is
provided by this horizontal component
which is n sine theta so what will be
saying in this case we will write the
equations show along vertical direction
so along vertical direction n cos theta
would be equal to mg and along the
horizontal direction
we have n sine theta is equal to the
centripetal force which is equal to MV
square by r let us suppose this is
equation 1 and this is equation now let
us suppose we divide equation 1 divided
by equation 2 so what do we get we get
caught theta is equal to mg divided by
MV square by R so this M and M gets
cancelled
so we get our G divided by v square so
this is the value of cot theta so now
here we can put the values which we
already have here now what would be the
value of V all we know the value of R
because this is given to PI R is equal
to thirty four point three so we know R
we know G now we need to calculate the
value of velocity now what would be the
Lassa T velocity would be the distance
traveled divided by the time taken so
the entire display distance traveled
here would be equal to thirty four point
three that is thirty four point three
meters divided by time taken is root
over twenty two right now we will put
this value in this equation so we find
cot theta would be equal to R what would
be our R would be thirty four point
three divided by two pi G would be nine
point eight and V Square would be thirty
four point three squared into twenty two
this is what we'll get on calculating
this we get the value comes out to be
one so we get the value of theta comes
out to be forty five degree that means
the cyclist will make an angle of 45
degree with the vertical so that he can
cover a part of this much circumference
in route over 22 seconds so I hope it is
clear to you now whenever you get a
question like this first of all you draw
the diagram so that you can visualize
the forces which are acting on the
particular body and then you write the
equation of motion in the horizontal as
well as vertical direction now let us
look at another problem the problem says
a circular racetrack of radius 300 meter
is banked at an angle of 15 degree if
the coefficient of friction between the
wheels of a race car and the
0.2 what is the optimum speed of the
race car to avoid wear and tear on its
tires so when will wear and tear happens
when there is no friction I already
discussed that when there is no friction
however the speed of the vehicle would
be not - at its maximum but there would
be wear and tear so if you don't want
wear and tear
then friction should be present right
now let us try to solve this let us see
what all values are given radius is
given as 300 meter the angle is theta is
given as 15 degree and the value of
coefficient of friction is given as zero
point 2 now we have already derived a
formula for the maximum velocity or the
optimum speed of the car to avoid wear
and tear that means the value of speed
after which wear and tear will start so
that speed is V naught which is V
maximum and what is the value of V
maximum it is RG time tan theta that
means the wear and tear will start only
when mu s becomes zero
so that means that value is the optimum
value of velocity so we will say this is
RG tan theta so let us put the values
this would be ad into G is 9.8 into tan
15 degree now putting the values we find
it as 28 point one meter per second now
let us look at the second part maximum
permissible speed to avoid slipping so
what would be the maximum permissible
speed to avoid slipping that means you
don't want the car to be slipped so in
that case it should be the scenario when
the value of the static friction is
equal to its maximum value that means
static friction should exist and the
value of static friction should be equal
to MU s into n this is that scenario
because when static friction is its
maximum value in that is it will not
slip slipping will start only when
becomes greater than this so add this in
this condition where do we get the value
of V the one which we derived for V Max
because we derived V Max considering F s
is equal to MU SL to N so V Max is equal
to what was the formula which we derived
it was our G into mu s plus tan theta
divided by 1 minus mu s into tan theta
now again here we put the values R is
equal to 300 G is 9.8 u.s. is 0.2 plus
tan 15 degree divided by 1 minus 0.2
into tan 15 degrees once you calculate
this you get the value as thirty eight
point one meter per second so most of
the times the questions which you get on
circular motion is where you need to
understand the understand what the
question is trying to say that is you
should have understood the concept that
when does a car slips or when does where
and clear start taking place so you
should know all these things all you
should understand this phenomena
logically so that you can apply the
correct formula in the correct place so
I hope this is clear to you now thank
you
visit exam viacom to watch free
educational videos try every online
tests get the best quality study
materials study from the best tutors and
mentors and much more thanks once again
motion part 25 is brought to you by exam
feel calm no more fear from exam please
make sure that you have watched all the
videos from part 1 part 24
before going ahead with power 25 start
with solving some new mannequins the
first problem says find the angle
through which a cyclist vents from the
vertical when he covers a circular path
of thirty four point three meter
self-confidence in route over 22 seconds
that means the self a cyclist is trying
to cover a circular part whose
circumference is given as thirty four
point three meters that means what is
the circumference of a circle it is 2 pi
add so value of 2 PI R is given as 34
point three meters and the time taken to
cover this much of distance is T is
equal to root over 22 seconds so
basically this 2 PI R is nothing but the
distance covered by the cyclist so but
through which angle he needs to bend
what is the angle that he makes with a
vertical so let us try to draw the
diagram so that he can able to visualize
the situation let us suppose this is my
vertical because it says that the
cyclist bends from the vertical let us
suppose this is the cyclist is just a
rough figure let us say this is the
cyclist like so he is making let us
suppose that he is making some angle
theta so we need to find the value of
this theta let us suppose he is making
some angle theta and he is trying to
take a curve now what will be the
different forces acting on the cyclist
because he is taking a turn somewhat
like this it's somewhat like this so
mass of the body will be in the I mean
weight of the body with in the downward
direction that is mg if M is the mass of
the cyclist now what what will be the
direction of the normal reaction normal
reaction is always by
when dick Euler to the surface of
contact now what is the surface of
contact head this is the surface of
contact if you see the cyclist is in
contact with the road at this point so
this is my surface of contact therefore
the normal reaction would be
perpendicular to this so this would be
the direction of the normal reaction
suppose if this is Theta so along this
direction what would be the component of
normal reaction this would be n cos
theta right so this would be n cos theta
therefore this side it would be n sine
theta correct now what is forcing the
cyclist to move in a circular path
obviously centripetal force centripetal
force would be along this direction
because it always act towards center
that means the centripetal force is
provided by this horizontal component
which is n sine theta so what will be
saying in this case we will write the
equations show along vertical direction
so along vertical direction n cos theta
would be equal to mg and along the
horizontal direction
we have n sine theta is equal to the
centripetal force which is equal to MV
square by r let us suppose this is
equation 1 and this is equation now let
us suppose we divide equation 1 divided
by equation 2 so what do we get we get
caught theta is equal to mg divided by
MV square by R so this M and M gets
cancelled
so we get our G divided by v square so
this is the value of cot theta so now
here we can put the values which we
already have here now what would be the
value of V all we know the value of R
because this is given to PI R is equal
to thirty four point three so we know R
we know G now we need to calculate the
value of velocity now what would be the
Lassa T velocity would be the distance
traveled divided by the time taken so
the entire display distance traveled
here would be equal to thirty four point
three that is thirty four point three
meters divided by time taken is root
over twenty two right now we will put
this value in this equation so we find
cot theta would be equal to R what would
be our R would be thirty four point
three divided by two pi G would be nine
point eight and V Square would be thirty
four point three squared into twenty two
this is what we'll get on calculating
this we get the value comes out to be
one so we get the value of theta comes
out to be forty five degree that means
the cyclist will make an angle of 45
degree with the vertical so that he can
cover a part of this much circumference
in route over 22 seconds so I hope it is
clear to you now whenever you get a
question like this first of all you draw
the diagram so that you can visualize
the forces which are acting on the
particular body and then you write the
equation of motion in the horizontal as
well as vertical direction now let us
look at another problem the problem says
a circular racetrack of radius 300 meter
is banked at an angle of 15 degree if
the coefficient of friction between the
wheels of a race car and the
0.2 what is the optimum speed of the
race car to avoid wear and tear on its
tires so when will wear and tear happens
when there is no friction I already
discussed that when there is no friction
however the speed of the vehicle would
be not - at its maximum but there would
be wear and tear so if you don't want
wear and tear
then friction should be present right
now let us try to solve this let us see
what all values are given radius is
given as 300 meter the angle is theta is
given as 15 degree and the value of
coefficient of friction is given as zero
point 2 now we have already derived a
formula for the maximum velocity or the
optimum speed of the car to avoid wear
and tear that means the value of speed
after which wear and tear will start so
that speed is V naught which is V
maximum and what is the value of V
maximum it is RG time tan theta that
means the wear and tear will start only
when mu s becomes zero
so that means that value is the optimum
value of velocity so we will say this is
RG tan theta so let us put the values
this would be ad into G is 9.8 into tan
15 degree now putting the values we find
it as 28 point one meter per second now
let us look at the second part maximum
permissible speed to avoid slipping so
what would be the maximum permissible
speed to avoid slipping that means you
don't want the car to be slipped so in
that case it should be the scenario when
the value of the static friction is
equal to its maximum value that means
static friction should exist and the
value of static friction should be equal
to MU s into n this is that scenario
because when static friction is its
maximum value in that is it will not
slip slipping will start only when
becomes greater than this so add this in
this condition where do we get the value
of V the one which we derived for V Max
because we derived V Max considering F s
is equal to MU SL to N so V Max is equal
to what was the formula which we derived
it was our G into mu s plus tan theta
divided by 1 minus mu s into tan theta
now again here we put the values R is
equal to 300 G is 9.8 u.s. is 0.2 plus
tan 15 degree divided by 1 minus 0.2
into tan 15 degrees once you calculate
this you get the value as thirty eight
point one meter per second so most of
the times the questions which you get on
circular motion is where you need to
understand the understand what the
question is trying to say that is you
should have understood the concept that
when does a car slips or when does where
and clear start taking place so you
should know all these things all you
should understand this phenomena
logically so that you can apply the
correct formula in the correct place so
I hope this is clear to you now thank
you
visit exam viacom to watch free
educational videos try every online
tests get the best quality study
materials study from the best tutors and
mentors and much more thanks once again
About this Video
Oct 04, 2025
Last updated
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