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Having described the kinetics of particles, we will discuss plane kinetics of rigid bodies.
We have not taken any lectures on the kinetics of system of particles, because I thought
that the same thing gets covered if I describe the kinetics of rigid bodies. A rigid body
is after all a system of particles in which continuous distribution of particles is there.
So, instead of algebraic additions, here you will get integrations. Actually, same thing
can describe the kinetics of system of particles, but instead of integration, you will have
discrete additions. Otherwise, essential thing remains same, procedure is entirely same.
So, therefore it gets covered here.
A rigid body is composed of several particles. Its equation of motion can be obtained by
applying Newton’s law to individual particles and integrating the combined effect. So, you
have got that. So, we have got a lot of particles here and they are of different this one.
We can take small particle of size dm, where dm is tending to 0; that means rigid body
is composed of infinite number of particles. However, its mass is finite, its individual
particle’s mass is 0. So, 0 multiplied by infinite can still give you a finite number.
That is how we do that. Therefore, we take a small particle of mass dm which is very
small differential element. The mass center is denoted by G and if we
take any axis system, like O x y z, then o times, O to G this distance is r bar and this
is O to dm that is r. So, this distance is r and then G to dm this distance is there.
Now, here we apply Newton’s law.
Let us consider a body, whose mass center G is located by the position vector r bar.
Then, r bar is actually rdm divided by m where this is r. We have written by bold letters;
that means it is a vector. So, rdm divided by m, m is the total mass of the body and
rdm, this is the mass equation of the mass center we have integrated.
This is based on the continuum hypothesis. We assume that all the particles, this whole
body is in a continuum and there are no voids in between the particles. That is our assumption
and we have been able to integrate.
Now, we apply Newton’s law for a particle of mass dm and then we have dF plus df is
equal to dm times r double dot. For that particle, the force acting on that particle is dF. It
is the resultant external force acting on the particle and df is the resultant internal
force acting on the particle. That is equal to dm times r dot dot, because r is the position
vector of that particle with respect to that outside inertial frame of reference. O is
some stationary point, r dot dot will give acceleration and you will have dF plus df
is equal to dm into r dot dot. Integrating, if we integrate both sides of
these equations, we integrate dF also. We integrate df and here we integrate dm times
r dot dot. Basically if we integrate dm r dot dot that is equal to m into r bar double
dot, where we have equation one. Equation one says that r bar is equal to integration
of rdm divided by m. So, we get dF plus integral df is equal to dm r dot dot; that is mass
times r bar dot, where r bar is the distance of the center of gravity from the fixed point
and m is the total mass of this thing.
If we integrate dF we get Fnet, where Fnet is net external force and if we integrate
df, small f then this will become equal to 0, since internal forces balance each other.
So basically, the internal forces will balance each other and therefore this is the thing.
Internal forces will balance each other. The resultant of the external forces acting on
the body equals the mass m of the body times the acceleration a of its mass center G. In
that case, you get only this equation that resultant of the external forces are mass
times the acceleration of the mass center.
It has to be noted that the acceleration of all particles in the body need not be same.
The body in this animation is moving with acceleration. So, you can see the animation.
This is undergoing rectilinear motion; that means this body is moving in a straight line.
In that case, all particles are moving with the same velocity and same acceleration. Therefore,
if we talk about the acceleration of mass center or of any other particle, that does
not matter.
However, in the other case, the acceleration in this animation, this disk is rotating.
The acceleration of all particles is different. Particle O is not moving. So, its acceleration
is 0; particle A is the having the maximum acceleration. Mass center is O. Although there
is acceleration of body; that means some particles are having acceleration, but overall, that
mass center is not having acceleration. Therefore, net resultant force, external resultant force
must be 0. That condition is there.
Let us talk about the angular momentum. The angular momentum of the mass system about
the mass center G is the sum of the moments of the linear momentum about G of all particles
and is given like this; suppose you have a fixed point O here and this is a mass center.
This distance OG is r bar, and from G to dm, I have denoted by Ri, this capital Ri. Then
small ri is the absolute displacement vector of dm with respect to O. You get HG is equal
to Ri cross dm Ri dot. So, this is the angular momentum about mass center, about G. But here,
I have taken the absolute velocity. So, dm; that means, you write the absolute linear
momentum and cross product it with Ri. that means distance from the mass center. Then,
you get expression for angular momentum.
HG is equal to Ri cross dm r dot i. If you differentiate this expression with time, then
you will get H dot G is equal to R doti cross dm r dot i plus Ri cross dm ri dot dot. We
applied the product rule of differentiation. It can be easily shown. It does not mean although
if the body is moving in a translator mode, then Ri remains in the same direction and
R doti is 0, because the distance between the particles of the rigid body cannot change.
However, if the body is doing rotation, in that case R doti need not be 0. This can be
very easily shown. Actually this first term, that means R doti cross dm Ri will be equal
to 0 and only the second term will remain. Now, this can be shown like this, that if
we can write Ri , this is like this. If this term Ri dot cross dm Ri dot is equal
to integral Ri dot cross dm, this will be R bar dot plus Ri dot. Like that we have decomposed.
The second term will obviously become 0, because cross product of R doti with R do ti is 0.
The first term becomes 0 because by the definition of the mass center. You have Ri dot dm can
be taken. Because it is a scalar quantity, you can take it like this. Then what happens,
you are left with R dot i cross dm and R dot.
This is 0, because we have that equation number one whereas here, it was shown that rdm. So,
if you can show that if you have taken this and differentiate this, obviously this will
come out to be 0. So, that way it can be shown. So, first term becomes 0 and the second terms
remains Ri cross dm into Ri dot.
Hence, we have H dot G is equal to Ri cross dm into Ri double dot. That is equal to basically
Ri cross dF plus df, because dF is the external force acting and this is internal force acting,
so dm times Ri dot dot. Here you can apply the Newton’s law and you get this thing.
These are the internal forces. So their moment will obviously be 0 and you are left with
Ri cross dF and that will equal to Mnet. Thus, the resultant moment about the mass center
of the external forces of the body equals the time rate of change of the angular momentum
of the body about the mass center. So, you have got H dot G is equal to Mnet.
We consider a body subjected to a number of forces. We can consider, it is subjected to
number of forces as shown in the free body diagram. Here, that we make the free body
diagram, in the free body diagram, we will show all the forces acting on the body. That
is called free body diagram. So, we see an arbitrary body with mass center G. We have
shown just four forces and one point has to be noted that these forces can be replaced
by an equivalent force and moment system. We can take the effect of all the forces here
and we can find out Fnet. That resultant is passing through G and then we have one couple
MGnet, because we discussed in the first lecture itself that the force system can be replaced
by one resultant force and then the net moment. So this is like that. Then we can obtain the
kinetic diagram. Kinetic diagram shows basically the inertia forces you can say that mass into
acceleration and mass into linear acceleration and mass into angular that moment of inertia
into angular acceleration. So, kinetic diagram is obtained by applying Newton’s law. Kinetic
diagram of the body is shown here.
We can see that net force is in some direction. In the same direction mass into acceleration
will be active then you have H dot G that is the angular momentum that angular momentum
is also shown and it is in the direction of MGnet, because we have established that equation
that rate of change of angular momentum will be in the direction of Mnet. So, you have
got here rate of change of angular momentum in this direction.
If we consider the plane motion in this slide, the mass center is shown as G and we choose
any other point, this point is O which is a fixed point. We take any other point P which
maybe on the body or it maybe on the hypothetical extension of the body. In that case, the distance
of P to G is R; that is this distance, the fixed distance. Then you have this d small
mass, dm that is at a distance of R dash from here. Then, we have G to dm is Ri. We can
attach axis system at G also and body maybe undergoing translation as well as rotation.
In that case what happens, the mass center G has an acceleration of a and the body has
an angular velocity omega is equal to omega k and angular acceleration alpha is equal
to alpha k, both taken in the positive z direction and so this will be alpha k. Because the z
direction of both omega and alpha remains perpendicular to the plane of this one, we
may use scalar notations omega and alpha to represent the angular velocity and angular
acceleration. We just can write these scalar equations. That will be enough.
In this case, if HG is equal to basically Ri, if we talk about HG is equal to Ri cross
dm, Ri cross dm R dot i plus r dot and this is Ri cross dm cross R dot i plus R cross
dm R dot. This term becomes 0, because this second term maybe written as R cross into
dmR with a minus sign which is 0, since the first moment of mass about the center of mass
is 0.
For a rigid body, the velocity of dm relative to G is R dot i is equal to omega cross Ri.
So, the magnitude of R dot i is omega Ri and it lies in the plane of motion normal to Ri.
The product Ri cross dmRi is then a vector normal to the x-y plane in the sense of omega
and its magnitude is Ri square into dm. Thus, magnitude of HG is equal to omega into Ri
square dm; that means, it is IG into omega, where IG is a constant property of the body.
It is measure of the rotational inertia or resistance to change in the rotational velocity
due to the radial distribution of mass about this thing.
Now MGnet is equal to HG that is IG times omega dot or this become IG into alpha. For
a rigid body in plane motion, you have two equations of motion. Fnet is equal to ma,
that is a vector equation. You can have two scalar equations, Fnet in x-direction is equal
to mass times acceleration in x direction and Fnet in y-direction is equal to mass times
acceleration in y direction, and MG net is equal to IG. So, these three equations are
enough to determine the motion in a plane.
We develop some alternative moment equation. You consider a point P different from center
of gravity. Angular momentum of the body about this point P is given by dHP is equal to Ri
prime into dm into r, because this distance is Ri prime from here. So, this is Ri prime
and that can be written as R plus Ri cross dm into r dot. So, this can be written as
this is equivalent to that.
Like, the first term maybe written as R cross dm into r dot is equal to r cross r dot dm
or R into mVG, where the VG is the velocity of the mass center. Second term is HG.
We make use of the principle of moment. So, we can also write MP net is equal to MG net
plus R cross Fnet. We know, in this case, MPnet will be H dot G and Fnet will be equal
to mass times acceleration.
Therefore, R is the vector from P to the mass center G, and a is the mass center acceleration.
We already know that H dot G is equal to I alpha and the cross product R cross ma is
simply the moment of magnitude m times a times d of ma about aP where d is the perpendicular
distance of acceleration vector from that point. Therefore, MPnet is equal to I alpha
plus mad. We can write, by parallel axis theorem I is
equal to IP minus mR square, because I is the moment, second moment of inertia about
the center of mass. Therefore, we can write like this and then it becomes MPnet is equal
to IP minus mR square into alpha plus ma times d and this becomes equal to IP times alpha
plus R cross.
If write that IP into alpha in the vector form, that means just put k, then after simplification,
this will be IP times alpha k plus R cross mass times acceleration. In this case, if
about any point P, P maybe on the body or it may be on hypothetical extension of the
body, you have the equation that MPnet is IP times alpha. Then you have got R cross
mass into acceleration product. If P is a fixed point, it is not moving. So,
this portion goes to 0. We have MPnet is equal to IP times alpha; that means, as you find
out the equation about the mass center, that moment about the mass center is MG net is
equal to IG times alpha. So, you get IP times alpha. So, that becomes valid. So, first condition
is that P should be stationary and then this is valid.
The second is that if aP and all are having the same direction, then also it is valid.
Thus, we say that this equation, if M is equal to general form that O is any point, and we
write M0 or MO equal to IO times alpha. In that case, this mass where MO is basically
the moment about O and IO is the mass moment of inertia about point O. So, this equation
is valid, provided three conditions; this O is mass center, O is same as G then also
it is valid. O is a fixed point or O is having one acceleration which is in the direction
of mass center, because R is the point having towards acceleration towards the mass center,
then also this equation is valid. Having discussed that kinetics of rigid bodies
in plane motion, now we describe the kinetics of body in different modes.
First, we decide that a rigid body in plane can move in two modes; one is the translation
other is rotation. Any motion can be broken down into two parts. One is the translatory
motion and other is rotary motion. In rigid body translation in plane motion, every line
in a translating body remains parallel to its original position at all times. That is
the definition. In rectilinear translation, all points move in straight lines, whereas
in curvilinear translation all points move on congruent curved paths. So, that means
you can have completely straight line type of motion or you can have a motion which will
be curved, but still that every line remains parallel to its original position and therefore
this can be called as this one.
I am showing one animation for rectilinear translation motion. Here, any particle of
the body is moving in a straight line. So, the particles here are moving in a straight
line. Similarly, the particles here are moving in another parallel straight line. So this
is the example of rectilinear translation.
This is a curvilinear translation. Here, if we follow any typical particle say mass center,
we will see that this mass center keeps on moving on the curved path. This is example
of curvilinear translation. However, during the motion, anytime if you see any line, this
particular edge of the body remains parallel to itself. So, it is not rotating. That angle
theta remains same. Therefore, omega, in this case is 0. So, translating body has angular
velocity 0 and angular acceleration is also equal to 0.
Therefore, for both translations omega is 0 and alpha is 0. So, for translation of a
body, Fnet then becomes equal to mass times acceleration, because that was one equation.
This is a vector equation, Fnet is a vector and that is equal to mass into a and MGnet
is equal to IG times alpha. So, MGnet is this one and so alpha is 0. Therefore MGnet is
0, means if you take about the mass center that MGnet will be equal to 0.
Let us discuss another type of motion; that means fixed axis rotation. Now, in the fixed
axis rotation, this is the body which is rotating about the point O. I have shown the mass center
G. Here, this is the mass center G and then you have got that at is equal to r times alpha.
It is rotating about this point. So, angular acceleration is given by alpha and the linear
tangential acceleration is equal to r times alpha, where r is the distance of O to G.
Similarly, the normal acceleration will be directed towards center that is equal to r
bar square, where omega is the angular velocity of rotation and alpha is the angular acceleration.
Now, in this body let us see what these equations are.
We consider noncentroidal rotation that means assume that O is not the center of mass. In
noncentroidal rotation, rigid body is constrained to rotate about fixed axis, not passing through
the mass center. For a rigid body rotating about a fixed axis O, all points in the body
describe circles about the rotation axis, at all times of the body in the plane of motion
and all lines of the body in the plane of motion have the same angular velocity and
angular acceleration alpha. This type of things we have discussed actually in the last classes.
The normal acceleration an is equal to omega square r bar and at is equal to r bar into
alpha. Therefore, these things will hold good. Then the equations of dynamics are basically
this. We get always these two equations; Fnet is equal to mass times acceleration and MG
net is equal to IG times alpha.
Thus, we have Fnet. Now, Fnet is equal to m alpha. This is basically a vector equation.
We can make out two scalar equations here one is the F normal net is equal to mr omega
square and F tangential net is equal to mr times alpha. Then, we get MGnet is equal to
IG alpha, which is of course scalar equation. Otherwise, we can write it as a vector equation
also. If we attach k bar, here k is the unit vector in the z direction. Since it is same,
actually both side we eliminate k and we write in a scalar form, MGnet is equal to IG times
alpha. These are the three equations we have got and by this, we can get the required things.
When you apply the moment equation about G, like you have to apply MG net is equal to
IG times alpha, then always remember that you have to account for the moments of the
all the forces applied to the body. That means, even you have to consider the forces, internal
forces which are applied at O. When you make a free body you have to understand that there
are some internal forces at O. These also have to be reaction forces and they have to
be taken into account. This force must not be omitted from the free
body diagram. Even if you omit some force which is passing thorough G, then maybe it
is alright, because when you take the moment about G that force in fact vanishes, but in
general you should show all the forces. Now, consider the free body diagram of the body.
For example, this body is acted by these forces and it was fixed at O. There must be some
reaction force acting. That force also has been taken into account here and this is r
and this is G. We make the equivalent kinetic diagram here. So, this is IG times alpha.
We show that this is IG alpha and then here mass times the tangential acceleration passing
through G and mass times that normal acceleration is basically m a n.
In kinetic diagram, we will be showing these things m a t and m a n and IG alpha. This
is a kinetic diagram. From free body diagram, one can easily obtain the kinetic diagram
by applying Newton’s law. The resultant force passing through G, you have to transform
these forces into a resultant force passing through G and then you can find out mass into
acceleration. Here, mass into acceleration vector maybe in the same direction as the
resultant force passing through G and then you can find out the IG into alpha, because
if you can find out the moment of all the forces about G, then you can get this thing.
We also have discussed in this lecture, this equation that if you take any point P, say
P is any point, then you get MPnet is equal to IP times alpha plus r cross mass into aP.
Here, IP is the moment of inertia of the body, second mass moment of inertia about the body.
But it is about point P, and this is r cross m aP here aP is the acceleration of point
P, not the point G, it is point P. So, we get this equation.
If we are talking about the fixed point O, then we can use this equation. Instead of
P just write O. You get MOnet is equal to IO times alpha plus r cross maO, However,
O is fixed. Thus, acceleration of aO is 0; so, a capital O is equal to 0. Therefore,
you get MOnet is equal to IO times alpha. So, this is the equation for the moment. For
the common case of rotation of a rigid body about a fixed axis through its mass center
G, if you have a rotation of a rigid body about a fixed axis which is passing through
the mass center G, then in that case a will clearly be 0 and therefore sigma F is equal
to 0. The resultant of the applied force is the couple IG times alpha. In that case, it
will be simply IG times alpha, because that is the thing. This way, these equations can
be used to study the rotation of a body.
We are going to introduce one concept that is called center of percussion. We will tell
about the center of percussion here. We have seen that if a body is rotating about fixed
point, not passing through its mass center, then the force system on the body maybe represented
by two forces passing through its center of mass in the normal and tangential direction,
together with a moment IG times alpha. You see always, in this case for example, you
know kinetic diagram has been shown. From kinetic diagram, one can obtain a resultant
free body diagram. Here, one can see the forces. Since in the kinetic diagram, you are having
that IG into alpha; that means, the moment should be present here. That moment is equal
to IG into alpha and mass times at is equal to Ft and mass times an is equal to Fn. Therefore,
what happens that if a body is rotating about fixed point not passing through the mass center,
then the force system on the body maybe represented by two forces passing through its center of
mass in the normal and tangential direction, together with a moment IG times alpha.
The moment maybe eliminated if the line of action of a tangential force is shifted to
pass from the point Q instead of G. In the same line you join O and G and then after
that you shift. So, as shown in the figure, the point Q is called the center of percussion.
I have made another diagram. In this diagram, the moment has eliminated; why because, there
was a force passing through G. I have shifted it in a parallel manner to another point Q
on the same line. So, this is, in effect if you take the moment about G you will still
get a moment, because the force is not passing through point G. So, there is no need to show
the force. Therefore, this has been shifted and then this point Q will be called center
of percussion. Let me repeat these points, because this is
very important. So, this point Q is a center of percussion, we have described. Now, from
another angle, for a general planar motion, we have these equations of motion that is
Fnet is equal to mass times acceleration and then MGnet is equal to IG times alpha. So,
these are basically three scalar equations Fnet is equal to mass times acceleration and
MGnet is equal to IG time is alpha.
Now, free body diagram and equivalent kinematic diagram are shown here. Free body diagram
shows the forces, externally applied forces and also the reaction forces passing through
O. G is the mass center and then the kinetic diagram is this; mass times tangential acceleration,
mass times normal acceleration is here and then you have got IG times alpha.
Consider a noncentroidal rotation about O. Like in this case, if we pass a tangential
force through G, if the tangential force was passing through G then mass times at will
be equal to F.
Where at is the tangential acceleration, the angular acceleration is given by alpha is
equal to at times r. Thus, you will have MGnet is equal to I times alpha. Since the force
F is passing through G, the moment of that force is about G0. Now, if it is from where
we do MG net then from where, because we have to balance this MG net is equal to I alpha,
obviously we get it from O, because there are some points at that O.
From the reaction at the pin you know that, because it is pinned at O we will get that.
Suppose a force F is applied at a point Q, which is at a distance of Q is k0 square by
r from O. Instead of that you know that is passing through G, if now we apply to pass
through Q, where k0 is the radius of gyration about O; that is, the moment of inertia about
O is given by I0 is equal to mk0 square, so this is given here.
Then you take the moment about O. You will have that Fq is equal to I0 times alpha, that
is the basic equation about O you have applied. So, alpha is equal to a F Fq by I0. We have
an expression for Q. So, it is Fk0 square divided by rmk0 square is equal to F divided
by mr and this is equal to F by m, a into1 by r; that means, this becomes equal to a
by r. Hence, the tangential acceleration is, a is equal to F by m. Therefore, the net tangential
force obviously must be F only. Thus, the force, the net tangential force is F and we
have already applied the force F passing through this one Q. Therefore, this force applied
by the pin must be 0. Therefore, we conclude that if a tangential force is applied at the
center of percussion, then no tangential reaction is developed at the fixed support.
We can also conclude that the sum of the moments of all forces about the center of percussion
will be 0, because if you take the moment about the center of percussion then that is
0. Therefore, if you want to get maximum advantage, sometimes you know batsman will hit the ball;
in that case, if he hits the ball at the center of percussion, he will not feel any reaction
at the hand. Therefore, that impact force will not be experienced by him and he can
still provide the angular momentum to the bat.
The center of percussion is very important in many situations to find out what is the
center of percussion. Now we discuss the other type of motion that is called rolling motion.
This is the motion of a disk or wheel rolling on a plane surface, the surface maybe horizontal
or it maybe inclined to this one. If the disk is constrained to roll without sliding, the
acceleration of its mass center and its angular acceleration are related; like, in the plane,
in pure rolling motion the acceleration of the mass center and the angular acceleration
will be related, but if there is some slippage taking place, then there is no relation. Now,
you can consider the different cases, like one, you can have balanced disk. A balanced
disk is that disk whose mass center and geometric center will coincide. If you have a cylinder,
then if its mass center is located at the center of the cylinder, then it is called
balanced. If mass center is different then you know that it is called unbalanced disk.
In that, the acceleration of mass center is angular acceleration times the radius, that
is radius of this one, because the body is in plane motion. The kinetic diagram of the
body consists of a horizontal force applied at the center and a couple. So, we can make
the kinetic diagram because body is in the plane motion.
Let us consider this case. Here, this is the plane surface and the cylinder is rotating.
This is the contact point C. From this, you are having a normal reaction N and then you
are having the mass F. Then, there is a weight W. So, this is G, that mass center is G and this is W. You have some frictional
force F. F actually will pass through this and then you may have some applied force acting
here and that is P and its weight passes through its mass center G. This is equal to the equivalent
kinetic diagram shown here. Here, the mass center is C and this is G that is passing
through this. So, you have mass times acceleration and this will be mass times r alpha and this
is I alpha. When a disk rolls without slipping, there
is no relative motion between the point of the disk in contact with the ground and the
ground itself. The point C is the instantaneous center of rotation. You get a friction force
F also here, but this force is self resisting; means, if you apply some force here P, the
same amount of resistance is developed here with a limiting value of F is equal to mus
times N. So, mus is the coefficient of friction into N.
When the disk rotates and slides at the same time, a relative motion exists between the
point of the disk which is in contact with the ground and the ground itself and the force
of friction has the magnitude. In this case, Fk is equal to muk times N, where muk will
be the coefficient of kinetic friction. Kinetic coefficient of kinetic friction is usually
less than the coefficient of sliding friction. Sometimes, it maybe as less as that; that
means, coefficient of kinetic friction maybe some 50% of the coefficient of sliding friction.
In other cases also, the difference may not be that much but there is some difference.
So, you get Fk is equal to muk times N. In this case, however, the motion of the mass
center G of the disk and the rotation of the disk about G are independent. You do not have
the formula, like V is equal to omega r. These types of things cannot be done. Similarly,
mass center acceleration cannot be said to be alpha times r; that will be less than or
equal to mus times N.
Either it will be less tangential friction force, either it will be less than mus times
N or it maybe at most equal; because, mus times N is the limiting value and s is the
coefficient of sliding friction. Then, you have one relation for geometry, that means
a is equal to r times alpha. So, you have, a is equal to r times alpha, where r is the
radius and alpha is the angular acceleration. Then you can have another situation that the
pure rolling is taking place, but sliding is impending; that means, sliding is about
to begin. In that case, F will be exactly equal to mus times N and you will have a is
equal to r times alpha, where alpha is the angular acceleration and r is the radius of
that disk. Then, you can have the third condition in
which there will be rolling but there will not be any sliding; both will go simultaneously.
In that case, the equation one will be that F is equal to muk times N that normal force
and tangential forces are related by kinetic coefficient of friction that is muk and a
and alpha are independent in this case. We will not have any relation between a and alpha.
When it is not known whether a disk is sliding or rolling, then what should we do? We should
first assume that the disk rolls without sliding. We assume that the disk is rolling without
sliding and pure rolling motion is taking place. If F is found smaller than or equal
to mus times N; that means, our assumption was correct.
If F is found larger than mus times N then the assumption of pure rolling was incorrect.
The problem should be started again. This time assuming that rolling and sliding. So,
the equation and also the expression for force will change and F will now become equal to
muk times N. Let us solve the problem on rolling.
This will be like this. I have shown here an inclined plane; this is at angle theta
and this is the metal hook G with a radius r that is released from the rest on the theta
inclined. If the coefficient of static and kinetic friction is mus and muk then we have
given that you determine the angular acceleration alpha of the hook and time t for the hook
to move the distance of S down the inclined. So, this is shown here in the figure.
If it is a hook and it is a hollow that type of thing here then moment of inertia about
G is given by mass times distance square radius square mra square. We may make use this relation.
If we make the free body diagram of this hook, you are having a force F, friction force which
is upward and then you will have a normal force N and then you have got a mass. Its
weight is passing through the center of gravity this is MG and then this is a and this is
alpha and this will be y. The counter clockwise angular acceleration requires, you have got
counter clockwise acceleration because this part, mass is coming down here. So, acceleration
is counter clockwise. So, you require a counter clockwise moment about G; that means F must
be upward. So, whatever direction we have shown here is correct, that means it is counter
clockwise.
Then, we make the kinetic diagram. In that, we just show that mass times acceleration
which is in the direction of inclined plane and that IG times alpha. Now, assume that
hook rolls without slipping, so that a is equal r times r alpha that equation is valid.
The three basic equations, equations of dynamics are written: one is that relating the tangential
force. So, mg sin theta is the force due to gravity minus F is equal to mass times acceleration.
Then you have got N minus mg cos theta is equal to 0; that is the normal direction.
The third is that moment, mg is equal to I alpha. Therefore, what happens this is F into
r is equal to mr square into alpha. That has been obtained by taking moment about the mass
center m. If we eliminate F between first and third,
F is appearing. We eliminate F, because we do not know this F. What is that? So, eliminating
that F, we get a is equal to g by 2 sin theta. This one we can put and of course N is equal
mg cos theta. That thing we are not using right now. So, we find out that eliminating
from this. We have eliminated this F and we put this. Then, we will be obtaining a is
equal to g by 2 sin theta and F will be equal to mg sin theta minus mg by 2 sin theta, that
means mg by 2 sin theta. From the second equation, we get N is equal
to mg cos theta. So, the limiting force is Fmax is equal to mus times mg cos theta, this
point is mg cos theta. We have to see that whatever force we have obtained that is mg
by 2 sin theta, should be more than mus into mg cos theta.
In case that you get Fmax is smaller than F then the assumption of pure rolling is wrong.
In this case, suppose you get that Fmax by that thing, it is mus mg cos theta and if
it comes out to be because this is the maximum possible value. In that case, a will not be
equal to r alpha and you will be getting F is equal to muk times N. You still have the
relation between the force and the normal reaction that is muk mg cos theta.
Now, you put this in the first equation, mg sin theta minus muk mg cos theta is equal
to mass times acceleration. In this case, the acceleration will be g sin theta minus
muk times g cos theta. You will get a different type of this one and we can of course use
that sigma mg is equal to IG alpha; that means, we can write muk times mg cos theta into r.
That will be equal to IG times alpha or alpha is equal to muk mg cos theta r divided by
IG. We can find out angular acceleration and linear acceleration, although they maybe unrelated.
The time required for the center G of the hook to move a distance S from the rest with
constant acceleration, obviously can be found by t is equal to under root 2 S by a, where
2 S is the distance because this is simple formula. S is equal to ut plus half at square
and this portion is being 0. We know, t is equal to 2S by a under root.
So, that means once we have found linear acceleration, be it in pure rolling motion or be it in rolling
*** sliding motion, we have found. Therefore, we can always find out time and that is under
root 2S by a and we can solve this problem. That way, by applying these basic equations
that you know, Fnet is equal to mass times the acceleration of the center that is the
vector equation. Also, M is equal to I times alpha and M0 is equal to I0 times alpha. By
applying these equations, we can solve any problem of kinetics in the plane motion.