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Lecture #7.2
Dynamics of Rotation
Last time we have started our discussion of dynamics of rotation and we saw that in
order for a body to have nonzero angular acceleration, it should be nonzero net torque
acting on the body.
We have introduced the concept of torque. In general case of a rigid body rotating

around fixed rotational axis some force F can 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 connecting it with the pivot point (rotational axis).
Then the torque of this force has the magnitude
  rF sin  ,
(7.2.1)


where  is the smaller angle between vectors r and F .
Torque obeys superposition principle. If there are several forces acting on the body,
one can find torques for each of those forces and calculate the net torque with respect to
the chosen rotational axis. The body can rotate around this axis with nonzero angular
acceleration if the net torque is not zero, in same way as the body moves with nonzero
acceleration only if the net force acting on it is not zero.
Let us now obtain mathematical equation for the Newton's second law in the case
of rotation. First we will do this for a small particle of mass m rotating around a circle of

radius r. Let F be the net force acting on this particle. This force can have two
components. For sure it should have a component towards the center of the circle, which
provides centripetal acceleration for this particle. It also may have tangential component
F , which is responsible for the angular acceleration and tangential acceleration.
Newton's second law for this component gives
F  mat ,
(7.2.2)
or multiplying by radius
b g
  rF  rmat  rm r ,
  I ,
(7.2.3)
where quantity I  mr 2 is known as the moment of inertia of the particle. Equation 7.2.3
has exactly the same form as the Newton's second law, if one replaces the net force by the
net torque, the acceleration by the angular acceleration and the mass by the moment of
inertia. So, the moment of inertia I is the angular analog of mass.
Equation 7.2.3 has been derived for a particle, but it can be generalized to the case
of a rigid body of arbitrary shape. Indeed any body can be considered as collection of
particles of different masses mi at different distances ri from rotational axis. One can
perform summation of equations 7.2.3 for those particles, which gives
    mi ri2 ,
i
i
(7.2.4)
  I ,
where angular acceleration is the same for all the particles of the rigid body, so it can be
taken out of summation. The net torque is the sum of all the torques for individual
particles and moment of inertia of the rigid body is
I   mi ri2 ,
(7.2.5)
i
where ri are the distances from rotational axis to different points of the body. If the body
consists of finite number of particles then equation 7.2.5 works. If, however, the rigid
body has continuous structure, calculations of moment of inertia become more
complicated and can be done on calculus basis only. The results of some of those
calculations are given in Table 9.1 on page 260 in the book.
Rolling
One of the most common examples of the rigid body’s rotation in 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 sliding) 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 vcm .
Point P of the wheel, where it contacts the ground should also move with speed vcm , so it
can stay exactly beyond point O. This means that for time t this point P moves along the
road for distance s  vcmt . At the same time, since this wheel moves without sliding, the
point of the wheel, which originally was on 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, so vcm 
s R

  R , so
t
t
vcm   R .
(7.2.6)
The linear velocity of wheel's rotation is the same as the linear velocity of the wheel's
translational motion. Every point of the wheel participates in both motions. The net
velocity of each point can be calculated as a vector sum of both velocities. The center of
the wheel O is not rotating. It only has velocity vcm of translational motion. The point at
the bottom of the wheel moves forward with vcm due to translation and backward with -
vcm due to rotation, so its net velocity is zero. The point T on the top of the wheel moves
forward with vcm due to rotation as well as due to translation, so its total velocity is 2 vcm .
Considering rolling of the wheel, I said 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 acm for its center of mass. In this case the wheel also accelerates and has the
angular acceleration, which can be related to acm as
acm 
vcm 

R R ,
t
t
(7.2.7)
where  is the angular acceleration of this wheel. For the wheel to have an angular
acceleration the nonzero net torque is needed. This torque comes from the force of
friction acting at point P. Since equations 7.2.6 and 7.2.7 are only valid in the case of the
smooth rolling without sliding, this means that the 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 if the
wheel is accelerating or slowing down.
Example 7.2.1. Find acceleration of the body rolling down the ramp.
The set of equations we obtained in the past for motion of the object along incline
plane is still valid
ma  mg sin   Ff ,
0  N  mg cos .
However, the force of friction in this equation is no longer force of kinetic friction. To

find this force we need to use Newton's second law for rotation. Force Ff 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 
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. Combing everything together we will find
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
.