Title: Physics 151: Lecture 21 Today
1Physics 151 Lecture 21Todays Agenda
- Topics
- Moments of Inertia Ch. 10.5
- Torque Ch. 10.6, 10.7
2Lecture 22, ACT 1Rotational Definitions
- Your goofy friend likes to talk in physics speak.
She sees a disk spinning and says ooh, look!
Theres a wheel with a negative w and with
antiparallel w and a !! - Which of the following is a true statement ?
(a) The wheel is spinning counter-clockwise and
slowing down. (b) The wheel is spinning
counter-clockwise and speeding up. (c) The wheel
is spinning clockwise and slowing down. (d) The
wheel is spinning clockwise and speeding up
3Example
See text 10.1
- A wheel rotates about a fixed axis with a
constant angular acceleration of 4.0 rad/s2. The
diameter of the wheel is 40 cm. What is the
linear speed of a point on the rim of this wheel
at an instant when that point has a total linear
acceleration with a magnitude of 1.2 m/s2? - a. 39 cm/s
- b. 42 cm/s
- c. 45 cm/s
- d. 35 cm/s
- e. 53 cm/s
4Moment of Inertia
See text 10.4
- Notice that the moment of inertia I depends on
the distribution of mass in the system. - The further the mass is from the rotation axis,
the bigger the moment of inertia. - For a given object, the moment of inertia will
depend on where we choose the rotation axis
(unlike the center of mass). - We will see that in rotational dynamics, the
moment of inertia I appears in the same way that
mass m does when we study linear dynamics !
5Parallel Axis Theorem
See text 10.5
- Suppose the moment of inertia of a solid object
of mass M about an axis through the center of
mass is known, ICM - The moment of inertia about an axis parallel to
this axis but a distance R away is given by - IPARALLEL ICM MR2
- So if we know ICM , it is easy to calculate the
moment of inertia about a parallel axis.
6Parallel Axis Theorem Example
See text 10.5
- Consider a thin uniform rod of mass M and length
D. Figure out the moment of inertia about an axis
through the end of the rod. - IPARALLEL ICM MD2
DL/2
M
CM
x
L
ICM
IEND
7Direction of Rotation
- In general, the rotation variables are vectors
(have a direction) - If the plane of rotation is in the x-y plane,
then the convention is - CCW rotation is in the z direction
-
- CW rotation is in the - z direction
8Direction of RotationThe Right Hand Rule
See text 10.1
- To figure out in which direction the rotation
vector points, curl the fingers of your right
hand the same way the object turns, and your
thumb will point in the direction of the rotation
vector ! - We normally pick the z-axis to be the rotation
axis as shown. - ??? ?z
- ?? ?z
- ?? ?z
- For simplicity we omit the subscripts unless
explicitly needed.
9Rotational DynamicsWhat makes it spin?
See text 10.6 and 10.7
video
- Suppose a force acts on a mass constrained to
move in a circle. Consider its acceleration in
the direction at some instant - a? ?r
- Now use Newtons 2nd Law in the ?direction
- F? ma? m?r
F
F?
a?
m
r
?
10Rotational DynamicsWhat makes it spin?
See text 10.6 and 10.7
- rF? mr2? use
- Define torque ? rF?.
- ? is the tangential force F?times the lever arm
r.
F
F?
a?
m
r
- Torque has a direction
- z if it tries to make the systemspin CCW.
- - z if it tries to make the systemspin CW.
-
?
11Rotational DynamicsWhat makes it spin?
See text 10.6 and 10.7
- So for a collection of many particles arranged
in a rigid configuration
m4
F1
m1
F4
r1
?
r4
m3
r2
r3
m2
F2
F3
12Rotational DynamicsWhat makes it spin?
See text 10.6 and 10.7
- ????????????????????? ?TOT
I???????????
- This is the rotational version of FTOT ma
- Torque is the rotational cousin of force
- The amount of twist provided by a force.
- Moment of inertia I is the rotational cousin of
mass. - 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 Nm.
13Torque
See text 10.6, 10.7, 11.2
- Recall the definition of torque
See Figure 10.13
14Example
See text 10.1
- You throw a Frisbee of mass m and radius r so
that it is spinning about a horizontal axis
perpendicular to the plane of the Frisbee.
Ignoring air resistance, the torque exerted about
its center of mass by gravity is - a. 0.
- b. mgr.
- c. 2mgr.
- d. a function of the angular velocity.
- e. small at first, then increasing as the
Frisbee loses the torque given it by your hand.
15Torque
See text 10.6 and 10.7
- ? r Fsin ?
- So if ? 0o, then ? 0
- And if ? 90o, then ? maximum
-
F
r
F
r
See Figure 10.13
16Lecture 21, Act 1Torque
- In which of the cases shown below is the torque
provided by the applied force about the rotation
axis biggest? In both cases the magnitude and
direction of the applied force is the same.
(a) case 1 (b) case 2 (c) same
L
F
F
L
axis
case 1
case 2
17Lecture 21, ACT 2
- A uniform rod of mass M 1.2kg and length L
0.80 m, lying on a frictionless horizontal plane,
is free to pivot about a vertical axis through
one end, as shown. If a force (F 5.0 N, q
40) acts as shown, what is the resulting angular
acceleration about the pivot point ? - a. 16 rad/s2
- b. 12 rad/s2
- c. 14 rad/s2
- d. 10 rad/s2
- e. 33 rad/s2
18Torque and the Right Hand Rule
See text 11.2
- 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 directionof the
torque.
F
y
DIRECTION t r X F MAGNITUDE t r F sin ?
r
x
?
z
19Torque the Cross Product
See text 11.2
- So we can define torque as
- ? r x F
- r F sin ?
- ?X y FZ - z FY
- ?Y z FX - x FZ
- ?Z x FY - y FX
F
f
?
r
20How Much WORK is Done ?
See text 10.8
- Consider the work done by a force F acting on an
object constrained to move around a fixed axis.
For an infinitesimal angular displacement d? - dW F.dr FRd?cos(?)
- FRd?cos(90-?)
- FRd?sin(?)
- FRsin(?) d?
- ? ?dW ?d?
- We can integrate this to find W ??
- Analogue of W F ?r
- W will be negative if ? and ? have opposite sign
!
?
F
?
R
drRd?
d?
axis
21Work Kinetic Energy
See text 10.8
- Recall the Work Kinetic-Energy Theorem ?K
WNET - This is true in general, and hence applies to
rotational motion as well as linear motion. - So for an object that rotates about a fixed axis
22Example Disk String
See text 10.8
- A massless string is wrapped 10 times around a
disk of mass M40 g and radius R10cm. The disk
is constrained to rotate without friction about a
fixed axis though its center. The string is
pulled with a force F10N until it has unwound.
(Assume the string does not slip, and that the
disk is initially not spinning). - How fast is the disk spinning after the string
has unwound?
See example 10.15
23Lecture 21, ACT 2
- Strings are wrapped around the circumference of
two solid disks and pulled with identical forces
for the same distance. Disk 1 has a bigger
radius, but both are made of identical material
(i.e. their density r M/V is the same). Both
disks rotate freely around axes though their
centers, and start at rest.
w2
Which disk has the biggest angular velocity
after the pull ? (a) disk 1 (b) disk 2 (c)
same
w1
F
F
24Example 2
- A rope is wrapped around the circumference of a
solid disk (R0.2m) of mass M10kg and an object
of mass m10 kg is attached to the end of the
rope 10m above the ground, as shown in the figure.
- How long will it take for the object to hit the
ground ? - What will be the velocity of the object when it
hits the ground ? - What is the tension on the cord ?
w
M
T
m
h 10 m
25Example Rotating Road
See text 10.8
- A uniform rod of length L0.5m and mass m1 kg is
free to rotate on a frictionless pin passing
through one end as in the Figure. The rod is
released from rest in the horizontal position.
What is - a) angular speed when it reaches the lowest
point ? - b) initial angular acceleration ?
- c) initial linear acceleration of its free end ?
L
m
See example 10.14
26Lecture 22, ACT 2
- A campus bird spots a member of an opposing
football team in an amusement park. The football
player is on a ride where he goes around at
angular velocity w at distance R from the center.
The bird flies in a horizontal circle above him.
Will a dropping the bird releases while flying
directly above the persons head hit him? - a. Yes, because it falls straight down.
- b. Yes, because it maintains the acceleration of
the bird as it falls. - c. No, because it falls straight down and will
land behind the person. - d. Yes, because it mainatins the angular velocity
of the bird as it falls. - e. No, because it maintains the tangential
velocity the bird had at the instant it started
falling.
27Example
See text 10.1
- A mass m 4.0 kg is connected, as shown, by a
light cord to a mass M 6.0 kg, which slides on
a smooth horizontal surface. The pulley rotates
about a frictionless axle and has a radius R
0.12 m and a moment of inertia I 0.090 kg m2.
The cord does not slip on the pulley. What is the
magnitude of the acceleration of m? - a. 2.4 m/s2
- b. 2.8 m/s2
- c. 3.2 m/s2
- d. 4.2 m/s2
- e. 1.7 m/s2
28Recap of todays lecture
- Chapter 10,
- Calculating moments of inertia
- Tourque
- Right Hand Rule