Take a look at this question here. This
is what we're going to be working
through today. "Two thin circular disks
of mass m and four m having radii of a
and 2a respectively are rigidly fixed by
a massless rigid rod of length L equals
square root of 24a through their
centers. This assembly is laid on a firm
and flat surface and set rolling without
slipping on the surface so that the
angular speed about the axis of the rod
is omega the angular momentum of the
entire assembly about the point O is L.
which of the following statements are true?"
This is a question from the 2016 JEE advanced exam paper two. Now this is an
exam that Indian high school students
would sit if they're wanting to get into
an engineering university now this
question was the most requested one for
me to go over and it's definitely
extremely difficult I think this
question is maybe intended to really
sort out who are the top students taking
this exam and many students may try to
skip over this one it's from the physics
section now physics is my area of
specialty so I was quite interested to
take a look at this note that in this
video I take a lot longer than the
probably recommended three minutes per
question to solve this particular one
some of that's because I'm explaining it
as I go but it actually is a really
difficult question and would take me I
think a lot longer than three minutes to
solve it especially in an exam situation
so it is very hard but with that being
said I think that you can get through it
and I hope that this video maybe makes
it a little bit clearer for you I'm
going to start by adding anything to the
diagram that we currently know or that
we can easily find out maybe through a
bit of trigonometry so we can assign
this angle in here to be theta and we
will be able to find sine theta and
cosine theta by enlarging this triangle
here in the corner finding sine and
cosine of theta will help us out later
on
a, l, this angle is theta, and in this
triangle this is the hypotenuse, so by
Pythagoras you can work it out to be the
square root of l squared plus a squared.
To find cosine of theta we have the
adjacent side and the hypotenuse, cosine
of theta is L over square root L squared
plus a squared. From the question we know
that L is equal to the square root of 24
times a so we can substitute that into
our equation for cosine of theta
when you complete the algebra that works
out to be the square root of 24 over
five. Likewise,
sine theta will be using the opposite
side and the hypotenuse and it will be a
over square root our squared plus a
squared and when you substitute in the
value for l here we end up with this
being one over five
let's also extend some of the lines on
our diagram I'm extending this line out
a bit and the reason for that is to mark
the direction of the angular velocity
vector when it is spinning around its
own axis now the angular velocity vector
you can find using your right hand and
wrapping your fingers around in the
direction of motion and the thumb will
point in the direction of that vector
this is one of the right hand rules of
physics and if we're spinning that way
then this is going to be our vector
pointing out this way now that's not the
only axis of rotation that we're dealing
with in this problem these wheels are
also moving around the origin and so we
want to mark that angular velocity
vector as well so if we're rotating like
this we sort of got our fingers on the
diagonal and wrapping them around in
that motion our angular velocity vector
is actually going to stick up a little
bit and I'll draw it on here this
angular velocity vector for the motion
of the wheels around the origin is going
to be offset from the z-axis by the
angle theta which is the same angle as
in here I'm going to mark that vector as
being Omega what we're also going to
want to do here before we really get
started is to find the location of the
center of mass of this whole assembly
we've got a pretty big mess here for M
and a smaller mess here so the center of
mass will be somewhere in between the
two but let's work out exactly where it
is now there is a formula to find the
location
the center-of-mass call that C ome and
you can find that by taking the M 1 X 1
plus M 2 X 2 over m 1 plus M 2 now M 1
and M 2 are just our two respective
masses X 1 and X 2 are the location of
these two masses from the origin and
that's all we're going to need to do so
let's put the numbers into there M 1
will be M its distance along this axis
is going to be L we're going to add that
to 4 m which is our second mass here
times its distance which is 2l that's
all going to be over m + 4 m 5 m
simplifying the algebra on this and
we're going to end up with nine fifths L
drawing miles a little bit
inconsistently I'll make them nice and
curly again okay so at nine fifths L
along this bar is going to be our center
of mass well 1 L will be five fifths so
at another four fifths along here is
going to be our center of mass so mark
that their center of mass and to just
make it nice and clear again and we're
talking about this angular velocity
vector here we want to know how far it
is from the horizontal axis and that
angle in there is going to be theta
because these two angles are opposite
each other
the last thing I want to do before we
really get started and answering the
question is to write down three more
formula which are reasonably simple but
will become very useful in answering the
question the first is that angular speed
is related to tangential speed or linear
speed and the radius by V of
I just know that every point on a
rotating object will have the same
angular speed but the tangential speed
will differ depending on the radius of
that point now we also want to know
something about angular momentum and if
we have that denoted by owl it is going
to be equal to the moment of inertia
times the angular velocity just refresh
your memory the moment of inertia of a
point mass is M squared where this is
the mass on a certain radius of rotation
and for a disc the moment of inertia is
M R squared over 2 where R is the radius
of the spinning disk first we want to
look at checking out the claims made in
part a of the question that the center
of mass of the assembly rotates around
the z axis with an angular speed of
omega over 5 now let's see if there's
any truth to that
so let's pick a point to study and we
are going to choose the center of this
first disc that point right in the
middle there and we want to find out its
linear velocity how it's moving through
space we already know its angular
velocity is little Omega and we can use
our trusty formula that we just spoke
about to find its linear velocity 2 so
let's have a look at that we can find
its linear velocity V from our Omega in
this case if we treat the bottom point
where it makes contact with the floor to
be stationary then that center point is
rotating at a radius of a so we can set
its radius to be a and therefore it's
linear speed to be a Omega that linear
speed is going to define how quickly
that point can move around in space
around the origin so we can now use that
to work out the magnitude of the angular
velocity vector about the origin that
capital Omega
will be
the linear speed divided by the radius
which is now the radius of going around
oh not just a little radius a and that's
going to be our linear speed we just
found a Omega over R which is going to
be owl in this case the distance from
the origin to the point that we're
talking about owl so that's all good we
found the angular velocity around o but
what the question was actually asking us
was to find the angular velocity around
zid so what we found is actually a
little bit offset from what we're trying
to find and it's offset by an amount
theta so if you know your trigonometry
and I can draw the triangle again
quickly here if this is Theta if this is
our z axis and this is our value we've
just found then since we have an
adjacent side and the hypotenuse these
two values can be related by Z is equal
to Omega time cosine theta we know from
what we worked out at the very start
that cosine theta is square root 24 over
five so if we put all our values in we
have a Omega over L times square root of
24 over five if we then also substitute
the value of L which we know to be
square root of 24 a we get a Omega
and if we cancel out some of the terms
you can get rid of these we are left
with Omega over five and that ladies and
gentlemen means that Part A of our exam
question that the angular speed around
the z axis is omega over five is
actually true so a is one of the correct
answers to this question
let's now look at Part B and we want to
see if the magnitude of the angular
momentum of the center of mass of the
assembly about the point O is this value
eighty-one ma squared Omega okay so
let's see if there's any truth to that
so what are we going to use for this
it's our trusty formula here that our
angular momentum can be found by
multiplying together our angular
velocity and moment of inertia now for
around the point O we already know what
our angular velocity is going to be and
all we need to know is find our moment
of inertia because we have a center of
mass on a massless rod rotating around
that point we can treat it as a point
mass and this here M R squared is the
formula for the moment of inertia that
we're going to use our angular momentum
around the capital Omega axis so note
that like that is going to be equal to
our moment of inertia there which is our
mass now what's our mass it's going to
be the center of mass and center of mass
means you can treat all of the mass of
the Assembly as being at that point so
it's going to be M plus 4 M it's going
to be 5 M rotating at a distance of 9/5
L because that was the location to our
center of mass so they're going to have
that's the moment of inertia times our
angular velocity which we worked out in
the previous part to be a Omega over L
nearly made a fatal flaw but don't
forget to square the radius there
M R squared Omega we're going to end up
with 5m 81 over 25 l squared we can
substitute L with its knowin value and
what we end up after doing that and
simplifying is 81 over 5 square root 24
a squared Omega M so let's go and have a
look at what the question was asking the
question claimed that it was 81 M a
squid Omega which is actually pretty
close because we had an 81 but it's not
exactly the right answer so for Part B I
would say no that is not a correct
possible answer you give it to them
though they do try to trick you if you'd
been rushing through this and seeing
that there was an 81 you might be like
yeah this is pretty close might go for
it but yeah pretty nasty they're trying
to trick you into giving one wrong
answer let's check out Part C now and
that's one thing us to find the
magnitude of angular momentum of the
assembly about its center of mass so no
longer about the point O but around the
axis that it's rotating on it claims
that this value is going to be 17 ma
squared Omega over 2 so let's see if
that's another lie or if that is true in
the previous part we found the angular
momentum along this axis and I call that
L capital Omega this time we're being
asked to find the angular momentum along
this axis here and I'll call that L with
a little omega the direction of these
quantities again comes from that right
hand rule we are rotating like this on
this axis so our angular momentum Victor
points down in this direction this time
to find the moment of inertia along this
axis we have two discs rotating on their
axis so we're going to use this formula
here that the inertia for a disc is M R
squared over 2 working this out on paper
we have our little Omega is going to be
the inertia of the first disc past the
moment of inertia of the second disc all
times Omega so inertia of the first disc
is going to be M times the radius that
it's rotating on which is going to be a
squared over two add that to 4m times
its radius which is 2a or squared that
over - that's our moment of inertia
terms and timing that by Omega working
through a little bit of the algebra here
and we're going to end up with 17 over
2ma squared Omega does that check out
with the question paper very nice to see
that actually yes it does this is the
same value here so another possible
correct answer I guess it sucks a few
quit at question a thinking you've got
the only correct answer cuz well there
was a second one waiting for you and we
haven't even tested the question D yet
so let's see if that's right or wrong
question D is asking us to find the
magnitude of the z component of the
total angular momentum it claims that
it's 55 ma squared Omega what we've done
in the previous two parts of this
question was find two different
components of the total angular momentum
now both of these previous values for
the angular momentum on each axis are on
diagonals so they have components along
the vertical and horizontal axes so we
just want to find the component of this
angular element which is in the vertical
direction and it looks like we're going
to be subtracting the component of this
which is in the vertical direction
because this one is pointing
downwards and this one is pointing
upwards from similar trigonometry that
we've already been using today to
convert this value into its vertical
component you will take this and times
it by cosine of theta and because you
might be able to see this one is in the
opposite direction we have an opposite
and our hypotenuse to find needed
component here we will take this value L
and times up by sine of theta so owl owl
zid component is going to be our capital
Omega times cosine theta subtract from
that our little Omega times sine of
theta this was our answer to be so we
can substitute that in here that 81 over
5 square root 24 a squared Omega M times
up by cosine of theta which we found out
right at the start
square root 24 over 5 subtract from that
the answer to Part C which was 17 over
2 m a squared Omega times it by sine of
theta which we also found out at the
start one-fifth now we're just going to
do a little bit of algebra here try to
simplify this as much as possible now
this to me seems like really
gastly algebra to try and simplify this
down without a calculator which I
believe you're not allowed in the exam I
don't know I guess we can keep going
doing more and more algebra to try and
get these numbers here to simplify out
to something and compare it to the value
of 55 which is what has claimed any
question paper me being a lazy little
person have glanced at this on a
calculator to see that these numbers
come out to be about 76 times M a
squared Omega which is not equal to our
proposed value of 55 so I'm going to say
that's a wrong
so but very big congratulations to
anyone who follows that through just by
pen and paper in fact congratulations to
anyone who attempted this question and
got to any point along the way it's
really you know it's a very advanced
question to have this in an exam it
requires lots of different steps of
thinking and you know I think this is a
really impressive exam question in my
last video when I reviewed the first
paper from the 2016 JEE exam I didn't
realize that there was a second paper
that also had to be solved on the same
day with very little break so knowing
that makes me respect people who have
set this exam even more because that's
very difficult to sit two exams you know
is so close to each other on the same
day. Thanks for watching this video and
please subscribe to my channel if you'd
like to see more videos like this, and a
very big good luck to anybody who is
currently studying for exams.
