Download Lecture #1, June 9

Lecture 5.1
Dynamics of Rotation
For some time now we have been discussing the laws of classical dynamics. However, for the
most part, we only talked about examples of translational motion. On the other hand, while studying
kinematics we have also learned about rotation. Let us look at dynamics of this process. In order to
do that, we will have to learn some additional properties of vectors.
1. Vector Multiplication
By now we have already discussed various physical quantities and we know that some
quantities can only have magnitudes and called scalars and other quantities may have both
magnitude and direction and these are called vectors.
We have also learned that the special steps have to be taken in order to add vectors. Both
magnitude and direction have to be taken into account when adding, so vector addition can not be
reduced to a simple algebraic addition.
Today we shall talk about multiplying vectors. Once again vector multiplication is very much
different from regular scalar multiplication.
There are two ways of how one can multiply vectors.


1) By definition, the scalar product of two vectors A and B (also known as the dot product) is
 
(5.1.1)
A  B  AB cos ,
where A and B are the magnitudes of these vectors and  is the angle between them. It is easy to see


that this quantity is a scalar, which has the significance of the projection of vector A on vector B or


the projection of vector B on vector A . It can be positive as well as negative depending on the sign


of the cosine-function. It reaches its maximum value, when vectors A and B are parallel, which
means that the angle between them is equal to zero and cosine of this angle equals to 1. The scalar
product becomes zero in the case of perpendicular vectors.
It is easy to see that commutative law works for the scalar product, which is
   
A B  B  A .
The scalar product can be expressed in terms of vectors' components as
 
A  B  Ax Bx  Ay By  Az Bz .
Exercise: Prove equation 5.1.3.
(5.1.2)
(5.1.3)
In addition to a scalar product of two vectors, one can introduce the mathematical object,
  
known as a vector product of two vectors. We shall call vector C  A  B to be the vector product


(or the cross product) of vectors A and B if it has the magnitude
C  AB sin  ,
(5.1.4)



where  is the smaller of the two angles between vectors A and B . The direction of this vector C
is defined according to the right-hand rule. We have already mentioned this rule when talking about


right-handed coordinate system. According to the rule, you have to place vectors A and B tail to
tail without altering their orientations and consider the line perpendicular to the plane formed by

these vectors. Place your right hand around this line in such a way that your fingers would sweep A

into B through the smaller angle  between them. Then your outstretched thumb points the direction

of C .
It is easy to see from this rule that commutative law does not work for the vector product,
instead the anti-commutative law takes place
 
 
B A   A B .
d
i
(5.1.5)


Another important property of the vector multiplication is that if vectors A and B are parallel or
anti-parallel, their vector product is equal to zero. The vector product has its maximum magnitude if


vectors A and B are perpendicular. In particular, if one considers the unit vectors along the
coordinate axes of the right-handed coordinate system, it gives
k  i  j ,
i  j  k,
(5.1.6)
j  k  i.
This means that the vector product of two vectors can be expressed in terms of the vector
components as
 
A  B  Ay Bz  By Az i  Az Bx  Bz Ax j  Ax By  Bx Ay k .
d
i b
Exercise: Prove equation 5.1.7, using equation 5.1.6.
g d
i
(5.1.7)
2. Torque and Newton's second law for rotation
Let us consider rotation of a body around the fixed axis of rotation. We have already seen that
the most effective way to describe kinematics of this motion is by means of the angular variables,
such as angular displacement, angular velocity and angular acceleration. Now we have to introduce
a similar set of variables in order to be able to describe dynamics of this motion. The main subject of
dynamics is force. Let us see what the angular analog of this quantity is.
In the case of translational motion, it does not matter at which point of the body you apply the
force as long as it is a rigid body which moves as a whole. Indeed if you consider a heavy box on the
floor, you can either pull it from the front or push it from behind with the same force and the result
is going to be the same. However, it is different in the case of rotation. For instance, you may notice
that a doorknob is always located as far from the door's hinge as possible. If you push this door right
near the hinge, you will never be able to open it. On the other hand even very heavy doors can be
opened without difficulty if you push them for the doorknobs. So, in the case of rotation it matters
not only what force you apply, but also how far from the rotational axis you apply it. Direction also
matters. If you push or pull this door in the direction along its surface it will not do any good for
opening of the door. You can open it only if your force has a component in the direction
perpendicular to the door.

In general case of a rigid body rotating around fixed rotational axis, some force F may be
applied in arbitrary direction at the point, which is located at distance r from the body's rotational

axis. The position of this point is defined by the radius vector, r , which connects the pivot point


(rotational axis) with that point. Then we shall define the torque  of the force F as the cross
product of two vectors
  
 r F.
(5.1.8)
The magnitude of this torque can be found as
  rF sin  ,
(5.1.9)




where  is the smaller angle between the vectors r and F . If vector r and force F form the plane

perpendicular to the rotational axis, torque  will be parallel or antiparallel to the rotational axis. In
a same way one can define the torque for a particle in a circular motion.

We said that force F is acting in the arbitrary direction, which means it may have a radial

component Fr along the vector r as well as the tangential component F  F sin  perpendicular to

the vector r . This means that
  rF  r F ,
(5.1.10)
where r  r sin  is the perpendicular distance between the rotational axis and an extended line

running through the vector F . This extended line is called line of action and r is called the moment

arm of the force F .
In our example about the door, we saw that if a force acts perpendicular to the door, it will be
most efficient way to open it. In this case torque of that force reaches its maximum magnitude. If the
force acts along the door it does not move the door, but in that case the torque of this force is equal
to zero. In a same way it is equal to zero if the force is applied directly to the hinge, because r=0 and
the door does not move either. So torque plays the same role for the rotation as force does for
translation.
The SI unit of torque is Newton-meter (Nm).
Since torque is a vector, it obeys the superposition principle. So if there are several forces
acting on a body, one can find torques for each of these forces and calculate the net torque with
respect to the chosen rotational axis. The body will rotate around this axis with non-zero angular
acceleration if the net torque is not zero in same way as the body will move with non-zero
acceleration only if the net force acting on it is not zero. On the other hand rotation with constant
angular speed does not require a non-zero net torque as it was in the case of the Newton's first law
for translation. However, it still requires a centripetal force, which has zero torque.
Now it is time to obtain mathematical equation for the Newton's second law in the case of
rotation. First, we will do so for the particle-like mass, m, rotating around the circle of radius r. Let

F be the net force acting on this particle. This force can have two components. It for sure should
have the normal component towards the center of the circle, which provides centripetal acceleration
for this particle. It may also have tangential component F , which is responsible for the angular
acceleration and tangential acceleration. Newton's second law for this component gives
F  mat ,
(5.1.11)
or multiplying by radius
b g
  rF  rmat  rm r ,
  I ,
(5.1.12)
where I  mr 2 is the moment of inertia for a particle. Equation 5.1.12 has exact same form as
Newton's second law, if one replaces the net force with the net torque, the acceleration with angular
acceleration and mass with moment of inertia. So, moment of inertia, I, is the angular analog of
mass. Equation 5.1.12 has been derived for a particle, but it can be generalized to the case of the
rigid body of arbitrary shape. Indeed, any body can be considered as combination of particles of
different masses mi at different distances ri from the rotational axis. One can perform the
summation of equations 5.1.12 for these particles, which gives


  i   mi ri2 ,
i


i
(5.1.13)
  I ,
where angular acceleration is the same for all of the particles in the rigid body, so it can be taken out
of summation. The net torque is a vector sum of all the torques for individual particles and moment
of inertia of the rigid body is
I   mi ri2 ,
(5.1.14)
i
where ri are the distances from the rotational axis to different points of the body. If a body consists
of the finite number of particles then equation 5.1.14 works. If, however, a rigid body has
continuous structure with infinite number of particles, the equation 5.1.14 has to be replaced with
the integral definition
z
I  r 2 dm ,
(5.1.15)
where integral is taken over the volume of the rigid body.
3. Calculating the moment of inertia.
The moment of inertia can be calculated according to equation 5.1.14 for a system of particles
and according to equation 5.1.15 for majority of the rigid bodies. The results for common bodies and
given rotational axes are presented in Table 10-2 in the book.
The question then becomes how the choice of rotational axis affects the moment of inertia.
Suppose we already know this quantity I com for the axis of rotation that extends through the body's
center of mass. This is the case for all the bodies in Table 10-2. In fact, we have not discussed the
definition of center of mass yet. We will do this in details later in the semester. Let me just briefly
mention that position for the center of mass of the rigid body is defined as

rdm

,
rcom 
m
z
(5.1.16)

where integral is taken over body's volume, m is the total mass of the body and r is the position
vector of the mass element dm. For all symmetrical bodies considered in Table 10-2, the position of
the center of mass is matched with geometrical centers of the bodies.
If we want to find the moment of inertia, I, for another axis, which is parallel to the original
axis going through the center of mass and the distance between two axes is h then
I  I com  mh2 .
(5.1.17)
The last equation is known as the parallel-axis theorem. To prove it, let us use the definition of the
moment of inertia 5.1.15. We shall place the origin of coordinate system to the body's center of mass
defined by equation 5.1.16. We will only need to consider the cross section of the body
perpendicular to both axes and going through the body's center of mass. Let this plane be the xy
plane of our coordinate system. We shall mark a and b to be the x, y coordinates of the new axis of
rotation at this plane. This means that according to Pythagorean theorem h2  a 2  b2 and
b g b g
2
r2  x  a  y  b
2
for any element dm having coordinates x and y, so
2
2
I   r 2 dm    x  a    y  b   dm    x 2  2 xa  a 2  y 2  2 yb  b 2  dm 


2
2
2
2
2
2
  x  y  2 xa  2 yb  h  dm   x  y dm  2a  xdm  2b  ydm  h  dm 
 I com  mh 2 ,




where we have made use of the fact that the origin was placed to the center of mass, so
I com 
zd
i
x 2  y 2 dm,
z
z
xdm  0 and ydm  0 .
4. Rolling
One of the most common examples of the rigid body’s rotation in the everyday life is rolling
of the bicycle's wheel or the car's wheel along the road. In fact rolling is not a pure rotation, because
the wheels are not only rotating but also performing translational motion along the surface of the
road. Let us consider this type of motion in details.
We shall consider a wheel rolling smoothly (without slipping) along the road. It participates in
both translational and rotational motions. Let the center of mass O (see the picture) of this wheel be
moving forward along the road with constant speed vcom .
Point P where the wheel contacts the ground should also move with speed vcom , so it can stay
exactly below point O. This means that for time t this point P moves along the road for a distance
s  vcomt . At the same time, since this wheel moves without slipping, the point of the wheel which
originally was at the ground rotates for the distance of the circular arc of the same length s  R .
Here  is the angle for which this wheel has turned and R is the radius of the wheel. Taking time
derivative of this equation one has
vcom  R .
b g
b g
d R
ds d vcomt
d

 vcom 

R  R , so
dt
dt
dt
dt
(5.1.18)
The linear velocity of wheel's rotation is the same as the linear velocity of wheel's translation. Every
point of the wheel participates in both motions. The net velocity of each point can be calculated as

the vector sum of both velocities. The center of the wheel O is not rotating; it only has velocity vcom

of translation. The point at the bottom of the wheel moves forward with vcom due to translation and

backward with - vcom due to rotation, so its net velocity is zero. The point T at the top of the wheel


moves forward with vcom due to rotation as well as due to translation, so its total velocity is 2 vcom .
Rolling can also be considered as pure rotation around the axis going through the point P
which will give the same results.
Considering rolling of the wheel, we assumed that it moves with constant speed. However, it
is not always the case. In fact, the car or the bicycle can accelerate along the road with acceleration

acom for its center of mass. In this case the wheel also accelerates and has the angular acceleration,

which can be related to acom by taking the derivative of equation 5.1.18 as
acom 
dvcom d

R  R ,
dt
dt
(5.1.19)
where  is the angular acceleration of this wheel. For a wheel to have angular acceleration the
nonzero net torque is needed. This torque comes from the force of friction acting at point P. Since
equations 5.1.18 and 5.1.19 are only valid in the case of smooth rolling without slipping, this means
that friction force acting on the wheel at point P is, in fact, the force of the static friction, not of
kinetic friction. This force opposes the wheel's tendency to slide. The direction of this frictional
force depends on whether the wheel is accelerating or slowing down. It is the same direction as the

direction of acceleration acom .
Example 5.1.1. Let us consider the situation similar to the example of the block sliding down
along the inclined plan. The only difference is that we will replace the block of mass m by the
rolling uniform body of the mass m. The question stays the same, to find acceleration of this body
rolling down the ramp.
The set of equations which we have obtained from the Newton's second law stays valid
ma  mg sin   Ff ,
0  N  mg cos .
(5.1.20)
However, the force of friction in these equations is no longer defined by the equation Ff  k N ,
because it is not kinetic friction any longer. To find this force we need to use Newton's second law

for rotation. Force F f is the only force in this problem, which has nonzero torque, because other
forces are applied either at the pivot point as gravitational force or along the radius as the normal
force, so we have
 f  I ,
RFf  I ,
Ff 
(5.1.21)
I Ia

,
R R2
where I is the moment of inertia for this body, R is the radius of the body and  is the angular
acceleration. So the last equation gives
ma  mg sin  
Ia
,
R2
FG m  I IJ a  mg sin ,
H RK
mg sin 
g sin 
a

FG m  I IJ 1  I
H R K mR
2
2
2
.