Download Rotational Motion Torque Moment of Inertia

Rotational Motion
Torque
Moment of Inertia
Lana Sheridan
De Anza College
Feb 29, 2015
Last time
• rotation and kinematics
• relating rotational and translational quantities
• torque
Overview
• torque
• moments of inertia
Torque
Torque is a measure of force-causing-rotation.
It is not a force, but it is related. It depends on a force vector and
its point of application relative to an axis of rotation.
Torque is given by:
τ=r×F
That is: the cross product between
• a vector r, the displacement of the point of application of the
force from the axis of rotation, and
• an the force vector F
10.4 Torqu
The component F sin f
tends to rotate the wrench
about an axis through O.
Torque
F sin f
S
F
r
f
O
S
d
Figure 10.7
r
F cos f
f
Line of
action
S
The force F has a
greater rotating tendency about an
axis through
O as
F increases
τ=r×
F=
rF sin φ and
n̂
as the moment arm d increases.
In our study of tra
we studied the cau
What is the cause
Imagine trying
ular to the door s
hinges. You will ac
force near the doo
When a force is
to rotate about th
axis is measured b
but we will consid
Chapter 11.
Consider the wr
perpendicular to t
where φ is the angle between r and F, and n̂ is the unit vector
perpendicular to r and F, as determined by the right-hand rule.
rom
F1, which has a moment arm
S
F2 is negative and equal to 2F 2d 2.
1
to rotate the object counterclockwise about an axis through O, and
S
F 2 tends to rotate it clockwise.
Example 10.3 - Net Torque on a Cylinder
A one-piece cylinder is shaped as shown, with a core section
protruding
the larger drum. The cylinder is free to rotate
es can cause
a changefrom
in translaw. Forcesabout
can alsothe
cause
a change
central
z axis shown in the drawing. A rope wrapped
e forces in causing this change
around the drum, which has radius R1 , exerts a force T1 to the
d the moment arms of the forces,
right
the cylinder. A rope wrapped around the core, which has
ts of force
timeson
length—newton
ed in these
units.RDo
confuse
radius
exerts
a force T2 downward on the cylinder.
2 , not
2 F 2d 2
re very different concepts.
What is the net torque acting on the cylinder about the rotation
axis)?
stubborn screw from a piece of
axis (which
is the
the z
d a screwdriver
for which
trying to loosen a stubborn
il, should you find a wrench for
y
S
T1
R1
inder
0.9, with a core section protrudrotate about the central z axis
he drum, which has radius R 1,
ope wrapped around the core,
n the cylinder.
R2
O
x
z
S
T2
Figure 10.9
(Example 10.3) A
ed in these units. Do not confuse
re very different concepts.
Example 10.3 - Net Torque on a Cylinder
stubborn screw from a piece of
d a screwdriver for which the
trying to loosen a stubborn
il, should you find a wrench for
y
S
T1
R1
inder
R2
0.9, with a core section protrudrotate about the central z axis
he drum, which has radius R 1,
ope wrapped around the core,
n the cylinder.
bout the rotation axis (which is
O
x
z
S
T2
Figure 10.9
(Example 10.3) A
solid cylinder pivoted about the z axis
S
through O. The moment arm of T1 is
S
R 1, and the moment arm of T2 is R 2 .
First: Find an expression for the net torque acting on the cylinder
about the continued
rotation axis.
Second: Let T1 = 5.0 N, R1 = 1.0 m, T2 = 15 N, and
R2 = 0.50 m. What is the net torque? Which way is the rotation?
Rotational Version of Newton’s Second Law
Tangential components of forces give rise to torques.
They also cause tangential accelerations. Consider the tangential
component of the net force, Fnet,t :
Fnet,t = m at
from Newton’s second law.
τnet = r × Fnet = r Fnet,t n̂
Now let’s specifically consider the case of a single particle, mass m,
at a fixed radius r .
The tangential force on the
10.5 Analysis M
in a torque onSecond
the
Rotational Versionparticle
of results
Newton’s
Law
particle about an axis through
A single particle, massthem,
at of
a the
fixed
radius r .
center
circle.
S
! Ft
m
S
! Fr
r
Figure 10.10
A particle rotating
tangential net force g Ft . A radial
S
net force g Fr also must be present
τ
=circular
rFnet,t
n̂
net
to maintain the
motion.
a circle
of a
For such a particle, in
Fnet,t
=under
matthe influence
S
= r m at n̂
In Chapter 5, we learned t
object and that the acceler
basis of the particle unde
is Newton’s second law. In
second law: the angular ac
proportional to the net to
complex case of rigid-obje
case of a particle moving i
ence of an external force.
Consider a particle of m
of a tangential net force g
The radial net force cause
etal acceleration. The tang
The magnitude of the net
dicular to the page throug
= r m (αr )
= (mr 2 ) α
Because the tangential ac
the relationship at 5 ra (E
Rotational Version of Newton’s Second Law
(mr 2 ) is just some constant for this particle and this axis of
rotation.
Let this constant be (scalar) I = mr 2 .
Rotational Version of Newton’s Second Law
(mr 2 ) is just some constant for this particle and this axis of
rotation.
Let this constant be (scalar) I = mr 2 .
Scalar I is not impulse! This is just an unfortunate notation
coincidence.
Rotational Version of Newton’s Second Law
(mr 2 ) is just some constant for this particle and this axis of
rotation.
Let this constant be (scalar) I = mr 2 .
Scalar I is not impulse! This is just an unfortunate notation
coincidence.
I is called the moment of inertia of this system, for this particular
axis of rotation.
Rotational Version of Newton’s Second Law
(mr 2 ) is just some constant for this particle and this axis of
rotation.
Let this constant be (scalar) I = mr 2 .
Scalar I is not impulse! This is just an unfortunate notation
coincidence.
I is called the moment of inertia of this system, for this particular
axis of rotation.
Replacing the constant quantity in our expression for τnet :
τnet = Iα
Rotational Version of Newton’s Second Law
Compare!
τnet = Iα
Fnet = ma
Now the moment of inertia, I, stands in for the inertial mass, m.
Rotational Version of Newton’s Second Law
Compare!
τnet = Iα
Fnet = ma
Now the moment of inertia, I, stands in for the inertial mass, m.
The moment of inertia measures the rotational inertia of an object,
just as mass is a measure of inertia.
Moment of Inertia
We just found that for a single particle, mass m, radius r ,
I = mr 2
However, this will not be the moment of inertia for an extended
object with mass distributed over varying distances from the
rotational axis.
For that case, the torque on each individual mass mi will be:
τi = mi ri2 α
And we sum over these torques to get the net torque. So, for a
collection of particles, masses mi at radiuses ri :
X
I=
mi ri2
i
Moment of Inertia
And we sum over these torques to get the net torque. So, for a
collection of particles, masses mi at radiuses ri :
I=
X
mi ri2
i
For a continuous distribution of mass, we must integrate over each
small mass ∆m:
Z
I = r 2 dm
Moment of Inertia
For a continuous distribution of mass, we must integrate over each
small mass ∆m:
Z
I = r 2 dm
For an object of density ρ:
Z
I = ρ r 2 dV
If ρ varies with position, it must stay inside the integral.
Moment of Inertia
Important caveat: Moment of inertia depends on the object’s
mass, shape, and the axis of rotation.
A single object will have different moments of inertia for different
axes of rotation.
Moment of Inertia
Important caveat: Moment of inertia depends on the object’s
mass, shape, and the axis of rotation.
A single object will have different moments of inertia for different
axes of rotation.
Also notice that these integrals and sums are similar to the
expression for the center-of-mass, but for I we have r 2 and we do
not divide by the total mass.
Units: kg m2
Moment of Inertia
For a collection of particles, masses mi at radiuses ri :
X
I=
mi ri2
i
For a continuous distribution of mass, we must integrate over each
small mass ∆m:
Z
I = r 2 dm
And for an object of density ρ:
Z
I = ρ r 2 dV
If ρ varies with position, it must stay inside the integral.
Example: Calculating Moment of Inertia
Calculate the moment of inertia of two equal point-like masses
connected by a light rod, length `, rotating about the center of
mass.
m
m
l
Example: Calculating Moment of Inertia
Calculate the moment of inertia of two equal point-like masses
connected by a light rod, length `, rotating about the center of
mass.
m
m
l
Let the CM the origin. It will be in the center of the rod.
Example: Calculating Moment of Inertia
m
m
l
Moment of inertia:
I =
X
mi xi2
i
2
2
`
`
= m −
+m
2
2
Example: Calculating Moment of Inertia
m
m
l
Moment of inertia:
I =
X
mi xi2
i
2
2
`
`
= m −
+m
2
2
=
m`2
2
n.
r5
m/V sometimes is
referred to as
Calculating
Moment
ofvolumetric
Inertiamass
ofdensity
a Uniform Rod
ass per unit volume. Often we use other ways of express10.7)
, when (Example
dealing with
a sheet of uniform thickness t, we can
ity s 5 rt, which represents mass per unit area. Finally, when
Moment
inertia of a uniform
rod of length
L and mass M
a rod of
uniformofcross-sectional
area A, thin
we sometimes
use
0 axis) and passing
about
an
axis
perpendicular
to
the
rod
(the
y
/L 5 rA, which is the mass per unit length.
through its center of mass.
ngth L and mass M (Fig.
and passing through its
y
y!
dx!
ure 10.15 (Example 10.7)
niform rigid rod of length L.
moment of inertia about the
xis is less than that about the y
. The latter axis is examined in
mple 10.9.
x!
O
x!
L
the definition of moment of inertia in Equation 10.20. As
Calculating Moment of Inertia of a Uniform Rod
Rod is uniform: let λ =
M
L
be the mass per unit length (density).
Z
Iy 0
=
r 2 dm
Calculating Moment of Inertia of a Uniform Rod
Rod is uniform: let λ =
M
L
be the mass per unit length (density).
Z
Iy 0
=
r 2 dm
Z L/2
=
−L/2
(x 0 )2 λ dx 0
Calculating Moment of Inertia of a Uniform Rod
Rod is uniform: let λ =
M
L
be the mass per unit length (density).
Z
Iy 0
=
r 2 dm
Z L/2
=
(x 0 )2 λ dx 0
−L/2
L/2
(x 0 )3
= λ
3 −L/2
M L3 L3
=
+
L 24 24
=
1
ML2
12
Summary
• applying kinematics
• torque
• moment of inertia
Next test Friday, Mar 4.
(Uncollected) Homework Serway & Jewett,
• Read ahead in Chapter 10.
• PREV: Ch 10, onward from page 288. Probs: 3, 7, 11, 15, 17,
19, 21, 25
• Ch 10, onward from page 288. Probs: 27, 29, 35, 39, 43