Transcript Physics 2211 Lecture 27

Physics 2211: Lecture 36

Rotational Dynamics and Torque
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 36, Page 1
Summary
(with comparison to 1-D kinematics)
Linear
Angular
a  constant
  constant
v  v 0  at
  0   t
s  s0  v 0t  21 at 2
   0  0t  21  t 2
v 2  v 02  2a  s  s0 
 2  02  2   0 
And for a point at a distance R from the rotation axis:
s = Rv = R
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
a = R
Lecture 36, Page 2
Rotation & Kinetic Energy

The kinetic energy of a rotating system looks similar to that
of a point particle:
Point Particle
K
1 2
mv
2
v is “linear” velocity
m is the mass.
Rotating System
1 2
K  I
2
 is angular velocity
I is the moment of inertia
about the rotation axis.
I   mi ri 2
i
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 36, Page 3
Rotational Dynamics

Suppose a force acts on a mass constrained to move in a
circle. Consider its acceleration in the ˆ direction at some
instant.

Newton’s 2nd Law in the ˆ direction:
a
F  ma

r F  mra
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
r
y
Multiply by r :
F
F

m
ˆ
x
Lecture 36, Page 4
rˆ
Rotational Dynamics


Define torque:   r F
 is the tangential force F
times the distance r.
Torque has a direction:
+ z if it tries to make the system
turn CCW.
- z if it tries to make the system
turn CW.
F
F
a
r
y

m
ˆ
x
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 36, Page 5
rˆ
Rotational Dynamics
  r F
 r  F sin    r sin  F
F
  r F
rp = “distance of closest approach”
or “lever arm”

F
Fr

r
rp
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 36, Page 6
Rotational Dynamics


 ri Fi   mi ri i
2
For a collection of many particles
arranged in a rigid configuration:
i
i
i
Since the particles are connected rigidly,
they all have the same .
2


m
r
 i  ii 
i
m4
i
I
F4
 NET  I
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm

r4
m3
F3
r1
F1
m1
r2
m2
r3
F2
Lecture 36, Page 7
Rotational Dynamics
 NET  I




This is the rotational analog
of FNET = ma
Torque is the rotational analog of force:
 The amount of “twist” provided by a force.
Moment of inertia I is the rotational analog of mass, i.e.,
“rotational inertial.”
 If I is big, more torque is required to achieve a given
angular acceleration.
Torque has units of kg m2/s2 = (kg m/s2) m = N-m.
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 36, Page 8
Comment on  = I

When we write  = I we are really talking about the z
component of a more general vector equation. (More on
this later.) We normally choose the z-axis to be the rotation
axis.)
z = Izz
z
Iz
z

We usually omit the
z subscript for simplicity.
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
z
Lecture 36, Page 9
Torque and the
Right Hand Rule

The right hand rule can tell you the direction of torque:
Point your hand along the direction from the axis to the
point where the force is applied.
Curl your fingers in the direction of the force.
Your thumb will point in the direction
of the torque.
F
y
x
r

Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
z
Lecture 36, Page 10
The Cross (or Vector) Product

We can describe the vectorial nature of torque in a compact
form by introducing the “cross product”.
The cross product of two vectors is a third vector:
B
AB  C


The length of C is given by:
C = AB sin 

A
C
The direction of C is perpendicular to the plane defined by A
and B and the “sense” of the direction is defined by the right
hand rule.
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 36, Page 11
The Cross Product

Cross product of unit vectors:
iˆ  i  jˆ  jˆ  kˆ  kˆ  0
iˆ  ˆj  kˆ
ˆj  iˆ  kˆ
kˆ  i  jˆ
iˆ  kˆ   jˆ
ˆj  kˆ  i
kˆ  ˆj  i
iˆ
+
+ 


+
kˆ
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
ˆj
Lecture 36, Page 12
The Cross Product

Cartesian components of the cross product:

 
C  A  B  Ax iˆ  Ay j  Az kˆ  Bx iˆ  By j  Bz kˆ

Cx  Ay Bz  By Az
B
Cy  Az Bx  Bz Ax

Cz  Ax By  Bx Ay
or
iˆ
ˆj
kˆ
C  Ax
Ay
Az
Bx
By
Bz
A
C
Note: B  A   A  B
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 36, Page 13
Torque & the Cross Product

So we can define torque as:
  r F
  r F sin
X = rY FZ - FY rZ =
Y = rZ FX - FZ rX =
Z = rX FY - FX rY =
F
y FZ - FY z
z FX - FZ x
x FY - FX y

r
y
z
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
x
Lecture 36, Page 14
Center of Mass Revisited

Define the Center of Mass (“average” position):
For a collection of N individual pointlike particles whose
masses and positions we know:
m2
N
RCM 
m r
m1
M
r1
i 1
i i
r2
RCM
m3
y
N
M   mi (total mass)
i 1
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
m4 r4
x
r3
(In this case, N = 4)
Lecture 36, Page 15
System of Particles: Center of Mass

The center of mass is where the system is balanced!
Building a mobile is an exercise in finding centers of
mass.
Therefore, the “center of mass” is the “center of gravity” of
an object.
m1
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
+
m2
+
m1
m2
Lecture 36, Page 16
System of Particles: Center of Mass

For a continuous solid, we have to do an integral.
dm
y
r
x
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
RCM
rdm  rdm



 dm M
where dm is an infinitesimal
element of mass.
Lecture 36, Page 17
System of Particles: Center of Mass

We find that the Center of Mass is at the “mass-weighted”
center of the object.
y
RCM
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
x
The location of the center
of mass is an intrinsic
property of the object!!
(it does not depend on where
you choose the origin or
coordinates when
calculating it).
Lecture 36, Page 18
System of Particles: Center of Mass



The center of mass (CM) of an object is where we can
freely pivot that object.
pivot
+
CM
Force of gravity acts on the object as though all the mass
were located at the CM of the object. (Proof coming up!)
If we pivot the object
somewhere else, it will
orient itself so that the
CM is directly below
the pivot.
This fact can be used to find
the CM of odd-shaped objects.
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
pivot
CM
pivot
+
CM

mg
Lecture 36, Page 19
System of Particles: Center of Mass

Hang the object from several pivots and see where the
vertical lines through each pivot intersect!
pivot
pivot
pivot
+
CM

The intersection point must be at the CM.
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 36, Page 20
Torque on a body in a uniform
gravitation field

What is the torque exerted by the force of gravity on a body
of total mass M about the origin?
M
y
dF  dm g
r
origin
dm
x
z
d  r  dF  r  dm g  rdm  g


1

   rdm  g   rdm  g    rdm   Mg  rCM  Mg
M

   r  dF  rCM  Mg
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 36, Page 21
Torque on a body in a uniform
gravitation field
M
y
origin
dm
dF
r
x
rCM
y
M
origin
x
rCM
z
z
   r  dF
  rCM  Mg
Mg
Equivalent Torques
Physics 2211 Spring 2005
© 2005 Dr. Bill Holm
Lecture 36, Page 22