physics 2211: lecture 36
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Physics 2211: Lecture 36. Rotational Dynamics and Torque. Linear. Angular. Summary (with comparison to 1-D kinematics). And for a point at a distance R from the rotation axis:. s = R v = R a = R . Rotation & Kinetic Energy. - PowerPoint PPT PresentationTRANSCRIPT
Lecture 36, Page 1 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Physics 2211: Lecture 36Physics 2211: Lecture 36
Rotational Dynamics and Torque
Lecture 36, Page 2 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Summary Summary (with comparison to 1-D kinematics)(with comparison to 1-D kinematics)
And for a point at a distance R from the rotation axis:
s = Rv = Ra = R
2 20 02
210 0 2s s v t at
2 20 02v v a s s
constantconstanta
AngularLinear
0v v at
210 0 2t t
0 t
Lecture 36, Page 3 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Rotation & Kinetic EnergyRotation & Kinetic Energy
The kinetic energy of a rotating system looks similar to that of a point particle:
Point ParticlePoint Particle Rotating System Rotating System
v is “linear” velocity
m is the mass.
is angular velocity
I is the moment of inertia
about the rotation axis.
21
2K I21
2K mv
2i i
i
I m r
Lecture 36, Page 4 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Suppose a force acts on a mass constrained to move in a
circle. Consider its acceleration in the direction at some
instant.
Multiply by r :
F ma
Fr mra
Newton’s 2nd Law in the direction:
m
F
F
a
rr
x
y
Rotational DynamicsRotational Dynamics
Lecture 36, Page 5 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Torque has a direction:+ z if it tries to make the system
turn CCW.- z if it tries to make the system
turn CW.
Define torque: is the tangential force F
times the distance r.
Fr
m
F
F
a
rr
x
y
Rotational DynamicsRotational Dynamics
Lecture 36, Page 6 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
rp = “distance of closest approach” or “lever arm”
FFr
r
F
sinr F sinr F Fr
Fr
rp
Rotational DynamicsRotational Dynamics
Lecture 36, Page 7 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
For a collection of many particles arranged in a rigid configuration:
2
i
i i i i ii i
r F m r
Since the particles are connected rigidly,they all have the same .
NET I
m4
m1
m2
m3
1r
4F
3F
2F
4r
3r
2r
1F
Rotational DynamicsRotational Dynamics
2
i i ii i
I
m r
Lecture 36, Page 8 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
This is the rotational analog of FNET = ma
Torque is the rotational analog of force:Torque is the rotational analog of force: The amount of “twist” provided by a force.
Moment of inertiaMoment of inertia I I is the rotational analog of mass, i.e.,is the rotational analog of mass, i.e.,
““rotational inertial.”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.
NET I
Rotational DynamicsRotational Dynamics
Lecture 36, Page 9 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
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 = Izz
We usually omit the
z subscript for simplicity.
Comment onComment on == II
z
z
z
Iz
Lecture 36, Page 10 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Torque and the Torque and the Right Hand RuleRight 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.
x
y
z
r
F
Lecture 36, Page 11 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
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:
The Cross (or Vector) ProductThe Cross (or Vector) Product
BA C
The direction of is perpendicular to the plane defined by and and the “sense” of the direction is defined by the right hand rule.
C
A
B
B
A
C
The length of is given by:
C = AB sin C
Lecture 36, Page 12 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Cross product of unit vectors:
The Cross ProductThe Cross Product
ˆ ˆ ˆ ˆ ˆ 0i i j j k k
ˆ ˆ ˆi j k ˆ ˆj k i
ˆ ˆk i j
ˆ ˆ ˆj i k ˆ ˆk j i
ˆ ˆ ˆi k j
i
jk
++
+
Lecture 36, Page 13 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
The Cross ProductThe Cross Product
Cartesian components of the cross product:
ˆ ˆ ˆ ˆxx y yz zA A A AB B B BC i j k i j k
B
A
C
Note:
AB A B
ˆ ˆ ˆ
x y z
x y z
A A A
i j k
B B B
C or
zx zyyC B ABA
xy xzzC B ABA
yz yxxC B ABA
Lecture 36, Page 14 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Torque & the Cross ProductTorque & the Cross Product
So we can define torque as:
x
y
z
r
F
sinFr
r F
X = rY FZ - FY rZ = y FZ - FY z
Y = rZ FX - FZ rX = z FX - FZ x
Z = rX FY - FX rY = x FY - FX y
Lecture 36, Page 15 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Center of Mass RevisitedCenter of Mass Revisited
Define the Center of Mass (“average” position):For a collection of N individual pointlike particles whose
masses and positions we know:
(In this case, N = 4)
y
x
m1
m4
m2
m3
1r
4r
2r
3r
CMR1
N
i ii
CM
m rR
M
1
N
ii
M m
(total mass)
Lecture 36, Page 16 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
System of Particles: Center of MassSystem 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
m2
+m1 m2
+
Lecture 36, Page 17 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
System of Particles: Center of MassSystem of Particles: Center of Mass
For a continuous solid, we have to do an integral.
y
x
dm
where dm is an infinitesimal
element of mass.
r
CM
rdm rdmR
Mdm
Lecture 36, Page 18 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
System of Particles: Center of MassSystem of Particles: Center of Mass
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).
y
x
We find that the Center of Mass is at the “mass-weighted” center of the object.
CMR
Lecture 36, Page 19 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
System of Particles: Center of MassSystem of Particles: Center of Mass
The center of mass (CM) of an object is where we can freely pivot that object.
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 objectsomewhere else, it willorient itself so that theCM is directly below the pivot.
This fact can be used to findthe CM of odd-shaped objects.
+ CM
pivot
+
CM
pivot
+
pivot
CM
mg
Lecture 36, Page 20 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
System of Particles: Center of MassSystem of Particles: Center of Mass
Hang the object from several pivots and see where the vertical lines through each pivot intersect!
pivot
pivotpivot
+
CM
The intersection point must be at the CM.
Lecture 36, Page 21 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Torque on a body in a uniform Torque on a body in a uniform gravitation fieldgravitation field
What is the torque exerted by the force of gravity on a body of total mass M about the origin?
d r dF r dm g rdm g
x
y
z
dm
dF dm g
M
origin r
1CMrdm g rdm g rdm Mg r Mg
M
CMr dF r Mg
Lecture 36, Page 22 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Torque on a body in a uniform Torque on a body in a uniform gravitation fieldgravitation field
x
y
z
dm
dF
M
origin r
r dF
CMr
x
y
zMg
Morigin
CMr
CMr Mg
Equivalent Torques