L24 : Circular motion on banked road - Laws of motion, Physics, Class 11 Video Lecture - Class 12

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please make sure that you have watched
all the weight rules from part 1 to 523
before going ahead with part 24 now let
us study the motion of a car on a banked
Road when I say banked Road I mean to
say that the road itself is constructed
in such a way that it is inclined at
some angle with the actual horizontal
let us suppose this is the actual
horizontal and this is angle theta so
this road itself is inclined at some
angle theta with the horizontal in the
previous slide we talked about level
road where the road was completely plain
from at when we compared it to the
normal horizontal but here you will
normally see these kind of roads when
you travel up hilly areas there you
would see that the outward of the roads
are generally inclined a bit up when
compared to the inner side of the road
so when the car when the car normality
moves along this kind of curved path so
we will study the motion for such parts
that is for such banked roads so let us
draw the diagram here let us suppose
this is the car of mass M it makes some
angle theta here so let us suppose the
weight which is acting downward would be
mg the normal reaction would act normal
to the surface of contact so this would
be n now if this angle is Theta we can
say this would be n cos theta so this is
Theta if this is n cos theta this would
be n sine theta sharing now there would
be a force of friction which would act
along this direction it has a we denoted
by small F now if this is small F and
this is Theta this entire thing is 90
degree so if this is 90 this is Theta
this will also become theta so this
angle will also become theta
so if this is f so this will be f cos
theta correct
now that we have drawn the Freebody
diagram for this vehicle which is moving
on a banked Road now let us write the
equation of motion let us write the
equation of motion in vertical direction
like let us suppose this is my y-axis
now this is my y-axis and this is my
x-axis correct so let us write it in the
vertical direction so in vertical
direction what would be the equation of
motion it would become now see if this
is f cos theta this will again become F
sine theta alright so in the vertical
direction it would be let us suppose
when the car if the car is at rest in
that case the in this direction it would
become n cos theta is equal to F sine
theta plus mg so that is n cos theta is
equal to F sine theta plus mg now in the
horizontal direction that is along this
direction we can say n sine theta plus F
cos theta is equal to this n sine theta
plus F cos theta in this case will
provide the centripetal force so we can
say this will be equal to F C that is
the centripetal force which is equal to
MV square by r in the previous case also
we saw that the force of friction
provided the centripetal force in this
case the force in this case a component
of the force of friction as well as a
component of the normal reaction both
together provide the centripetal force
which makes the car move in a curved
path so we get these two equations of
motion now let us try to solve the two
equations now we already know that F is
less than equal to MU s into n that is
Fric force of friction that is FS is
less than equal to MU s into n so for
velocity to start that means for the
body to move the value of F should be
its maximum
and the maximum value is mu s into n now
we will put this value of f in the above
two equations so what do we get we get n
cos theta is equal to MU s n sine theta
plus mg and then n sine theta plus mu s
n cos theta is equal to mV square by r
now let us mark this equation as one let
us say this equation as two now from the
first equation let us from Equation 1 we
can say n cos theta just note that
although the equation seems to be very
big but it is nothing but simple
mathematical calculations so this is
equal to mg so from Equation 1 what we
are trying to do we are trying to write
n in terms of all other terms so this
will be equal to MU s divided by cos
theta minus mu s sine theta so this is
what we get the value of n now we put
this value of n in equation 2 when we
put this value of n in equation 2 then
what do we get get mg sine theta divided
by cos theta minus mu s sine theta plus
mu is n cos theta I am sorry
here also you will have to put the value
mu s instead of n it would be mg divided
by cos theta minus mu is sine theta this
is equal to mV square by r now we will
solve it mg sine theta plus mu s cos
theta divided by cos theta
- view s sine-theta this is equal to mV
square by R so we found mg sine theta
plus mu s cos theta divided by cos theta
minus mu s sine theta is equal to mV
square by R now to simplify the
calculations let us divide the numerator
as well as denominator by cos theta so
what do we get we get mg tan theta plus
mu s divided by 1 minus mu s tan theta
this is equal to mV square by R I'm
sorry this is smaller so M M gets
cancelled so we can say v square is
equal to RG tan theta plus mu s divided
by 1 minus mu s tan theta or we can say
the maximum velocity is equal to root
over R - s plus tan theta divided by 1
minus mu is tan theta so here we see the
maximum velocity of the vehicle in a
circular path on a banked road would be
given by this value so in the previous
slide we saw the maximum velocity that a
vehicle can have
in a circular path when it is moving on
a level road so in this is this is the
value of maximum velocity if the vehicle
is moving on a banked Road
why is it the maximum velocity because
if you go back to the previous slide and
check if you go back to the beginning of
this calculation itself you will see
that we considered that if the velocity
when the velocity will be maximum in
that case the friction the value of
static friction will be equal to mus
into n so we started deriving this
relation taking into consideration that
the value of static friction is its
maximum value so when value of static
friction would be the maximum value then
the velocity will also be maximum so
that is how we get the value of maximum
velocity to be this much now we will go
ahead and have a comparison between the
two cases so if you compare this value
with the previous value you will find
that the velocity is greater for bank
loans than level roads that is why in
many of the areas the roads are
constructed slightly back so that the
velocities of the cars while taking the
turn is more so we find that the valid
value of V Max in case of banked roads
is greater than the value of V Max in
case of level roads now we will look at
a special case of motion of car on a
banked Road so what is that special case
let us say if the value of mu is in the
in the expression which we derived just
now for V Max let us say if mu X mu s is
equal to 0 nu is equal to 0 would mean
that there is no friction right so if
there is no friction then what would be
the value of v max if you put the value
of MU is to be 0 now what we derived we
derived the formula to be V Max is equal
to ad into G mu s plus tan theta divided
by 1 minus mu s tank
down now if mu is is equal to zero in
that case this value would become root
over ad G tan theta this would become
the value of maximum velocity so what
does we convert were to be infer from
this we mean to say that when speed is
v-max that is no frictional force that
means when there is no frictional force
in that case the maximum velocity of the
vehicle is equal to this much so in this
case the centripetal force is given only
by the normal reaction only by the form
component of normal reaction and not by
the frictional force so when the
milettis a we denote this value of v-max
as V naught so when the velocity of the
car is V naught in that case no
frictional force is needed to provide
the centripetal force so who will
provide the centripetal force
centripetal force will be provided by
the component of the normal reaction
which component of normal reaction that
is n sine theta
go back to the derivation and field see
the first relation you would come to
know the centripetal force would be
provided by then come this component of
normal reaction however when this is the
velocity that is when there is no
friction at all and in those cases
little where and clear of tires take
things so that is one kind of
disadvantage so what did we observed by
studying the motion of a car on a level
road as well as banne throat we found
that the maximum velocity that a car can
have on a circular path is more in case
of a banked road when compared to a
level road we also found that in such
circular motion the centripetal force is
the one which is responsible for the car
to move in circular motion and the
centripetal force is provided by the
force of friction in case of level roads
and it is provided by centripetal and it
is provided by the component of normal
reaction and component of frictional
force in case of banked roads before
that let us have a comparison between
motion along
level and Bank as I already told the
maximum velocity of a car in circular
motion is greater on banked roads than
on level roads thank you
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