Download Mechanics 105 chapter 10

Rotational motion (chapter
ten)

Angular speed, acceleration
 Rotational kinematics
 Relation between rotational and
translational quantities
 Rotational kinetic energy
 Torque
 Rigid body in equilibrium, under a net
torque
 Angular momentum and its conservation
 Precession
 Rolling of rigid bodies
Angular speed
Relation between linear and
angular displacement
  sr
s
r

The angle here must be measured
in radians
(conversion:  radians = 180°)
The average angular speed can be
defined from the angular displacement
in some time interval as
And the instantaneous angular speed as
Units: rad/s (rad is not a “real” unit)
 
 f i
t f  ti
d

dt
Angular acceleration
Likewise, we can define average and
instantaneous accelerations as
 
 f  i
t f  ti
d

dt
For a rigid body (points remain fixed with
respect to each other) all parts have same
angular speed and acceleration
These quantities are vectors – their direction
is found using the right hand rule
Rotational kinematics
Using the same derivations as we used to find
the equations of linear kinematics, we find
 f   i  t
1 2
 f   i   i t  t
2
 2f   i2  2 ( f   i )
1
 f   i  ( f   i )t
2
For constant
angular
acceleration
Relation between rotational and
translational quantities
The angular kinematic variables will be
related to the linear ones for points on a
rigid body according to:
ds
d
Note this is for
v
r
 r
dt
dt
the tangential
d 2s
d
acceleration
at  2  r
 r
dt
dt
We also know the
v2
expression for the
ar 
 r 2
r
centripetal acceleration
Therefore, the total acceleration vector at
any instant is

a  ar rˆ  at tˆ  r 2 rˆ  rtˆ

a  ar2  at2  r  2   4
Note: rotational variables depend on
choice of axes (location)
ConcepTest
Examples

A ball of glass will bounce higher than a ball of rubber.
 Thomas Edison was afraid of the dark.
 A U.S. Quarter has 119 grooves on its circumference.
A dime has 118 grooves.
 It is impossible to lick your elbow.
 It is physically impossible for pigs to look up into the
sky.
 Over 75% of people who read this will try to lick their
elbow.
Rotational kinetic energy
The total kinetic energy of an extended object can be
found by adding up the energy of its constituents.
For a collection of discrete particles:
1
1
2
K   K i   mi vi   mi ri 2 i2
2 i
i
i 2
but since all  i  
1
1 2
2 2
K R    mi ri   I
2 i
2

Where I is a quantity named the moment of inertia
Rotational kinetic energy
The resulting equation has a similar form to the one for
the linear (translational) kinetic energy
Note that this is not a new type of kinetic energy – only an
expression for the translation KE for a rotating rigid body
For a continuous object
I   r 2 dm   r 2 dV
where  is the density and V the volume, i.e., the second
integral is over the spatial dimensions of the object.
ConcepTest
Examples
Moments of inertia of common objects
Solid cylinder, radius R: I=½MR2
Hollow cylinder, radii R1 and R2: I=½M(R12+R22)
Thin cylindrical shell, radius R: I=½MR2
Solid sphere: (2/5)MR2
Thin spherical shell: (2/3) MR2
Thin rod, length L, about center: ML2/12
Thin rod, length L, about end: ML2/3
Torque
Torque is the rotational quantity corresponding to force – it
represents the tendency of a force to cause an object to rotate.
The expression for torque is:
 
  r F

This is a vector product. The magnitude of the vector product is
 
  r F sin 

The direction of the vector product
is perpendicular to the plane
formed by the two vectors forming
the product, in a direction given
using the right hand rule.



r


F
Torque
We can think about the magnitude in two ways. One is that Fsin
is the component of the force perpendicular to the position r. The
second is that rsin (the “moment arm”) is the perpendicular
distance from the rotation axis to the line of action of the force.
For extended objects, equilibrium implies both zero acceleration
and zero angular acceleration. Since torque is the tendency of a
force to cause rotation, the net torque on an object in equilibrium
must be zero

  0
ConcepTest, Example
about any axis!!!
Rigid body under a net torque
If the torque on a rigid object about any axis is not zero, the points
making up the object will be experiencing a net force, and so
accelerate.
The tangential acceleration of the ith mass will be governed by
Fti  mi ati
Multiplying by the position gives
Summing over all particles gives
ri Fti  mi ati ri
 i  mi i ri 2
   i   mi ri 2  i  I
i
i
Rigid body under a net torque
Example: Atwood machine with a massive
frictionless pulley
+y

F

T1
m1

m1 g
m1
m2
F
F
I

T1

T2

T2

mP g
1
 T1  m1 g  m1a
2
 m2 g  T2  m2 a
1
2 a
  T2 rp  T1rp  I   2 M p rp  r
p
m2

m2 g
3 unknowns:
T1, T2, and a
Rigid body under a net torque
T1  m1a  m1 g
T2  m2 g  m2 a
1
T2 rp  T1rp  (m1a  m1 g  m2 g  m2 a)rp  M p rp a
2
(m1  m2 ) g
a
(m1  m2  M p / 2)
Work and energy in rotational motion
The work done by an external force on a point P
on a rigid rotating body is
 
dW  F  ds  ( F sin  )rd  d
Where  is the angle between the direction of
the force and the position of its application,
and d is the infinitesimal angle over which
the object rotates.
Examples
Work and energy in rotational motion
If we rewrite the expression for torque as
d
d d
d
  I  I
I
I

dt
d dt
d
d  Id
Integrating over  between an initial and final
angular speeds, we find the rotational version
of the work-kinetic energy theorem
f
f
1 2 1 2
W   d   Id  I f  I i
2
2
i
i




Donkeys kill more people annually than plane
crashes.
23% of all photocopier faults worldwide are caused
by people sitting on them and photocopying their
buttocks.
Walt Disney was afraid of mice.
The average human eats eight bugs or spiders in
their lifetime while sleeping.
Angular momentum
Each point on a rotating object has a velocity and a mass, and
so even if the object does not move (translate) it has a
momentum associated with it.
We call this the angular momentum, defined as
  
Lrp
The vector angular momentum is a cross product
of the radius and the linear momentum
Note that it depends on the choice of axis!
Also note that the cross product is not
 
commutative:   
L  r  p  pr
Angular momentum
Rewriting the total angular momentum of a system of
particles as
L   mi vi ri   mi riri  ( mi ri 2 )  I
i
i
i
We can see that the angular momentum is the analog of the
linear momentum in equations of dynamics




dL d  
dr   dp  dp
 (r  p) 
 pr
r
dt dt
dt
dt
dt This the rotational


version of Newton’s
 
 dp
  r F  r 
nd law
2
dt

 dL
 
dt
Angular momentum
Likewise, for a system of particles

ext


 dLtot
dLi d

  Li 
dt dt i
dt
i
This means that the angular momentum of a system
about a specific axis will be constant if the external
torque about that axis is zero
For an isolated system, then
L  I  constant
ConcepTest
Examples
Precession
A top spinning on a horizontal surface
with angular speed  will have an
angular momentum L with magnitude I,
pointing in a direction along the axis of
rotation
If it makes some angle with the vertical,
the gravitational force will give a torque
about the point of contact with the
surface.
This torque will cause the angular
momentum to change direction, tracing
out a circle perpendicular to the vertical
direction – this behavior is called
precession.
From Serway & Jewett, Principles
of Physics, 3rd ed., Harcourt 2002
Rolling motion
As a cylinder or sphere or radius R rolls through an
angle , its center of mass moves s=R , so
vCOM
aCOM
d
R
 R
dt
d
R
 R
dt
But note that the object is not rotating about its center –
the bottom is stationary at any point, while the top
moves at a speed 2vCOM
Rolling motion
We can write the total kinetic energy of the object as
1
1
2
2
K  I COM   MvCOM
2
2
Parallel axis theorem: The moment of inertia through any
axis parallel to the axis through the COM follows the
following relation:
I p  I com  MD 2
where D is the distance between the axes
Rolling motion
Using the parallel axis theorem for a cylinder, we see
that
1
1
1
1
2
2
2
K  I COM   MvCOM  I COM   MR 2 2
2
2
2
2
1
1
2
2
 I COM  MR   I p 2
2
2


In other words, the kinetic energy is equal to that of a
purely rotating cylinder about the point of contact with
the ground.
where D is the distance between the axes