P221 Exam 2 Histogram
Average: 45.0/78 (57.6%)
6
5
Frequency
4
Frequency
3
2
1
0
18
24
30
36
42
48
54
Bin
60
66
72
78
Vector Nature of w
Torque as a Vector Cross Product
t = r x F |t| = rFsinq
The direction of the
torque (t) is that of an
advancing RH screw
turned from the
direction of r toward the
direction of F
See: http://www.phy.syr.edu/courses/java-suite/crosspro.html
Basics of Rotational Dynamics
I=
2
Si miri
K=½I
2
w
t = r x F |t| = rFsinq
t=Ia
Rotational Inertia for Selected objects and rotation axes: HR&W Table 10-2
Chapter 10 problems
1
2 h
h=0
CALM Q
Hoop 1
Ball 2
Hoop rolls without slipping, ball drops.
Each m=M
• [time: Ball first (13); hoop (1) same (5): Speed ball (7)
hoop (9) same (2); KE ball (3), hoop (8) same (6)]
• The ball arrives first, the hoop is moving faster just before it reaches
height = 0, the hoop has greater kinetic energy just before it reaches
height = 0. [This was the most common answer]
• ball hits first... ball is faster... ball has more KE... all due to friction on
hoop... [This got two of the three parts correct, but misses the key
point about kinetic energy, so maybe the speed was fortuitous?]
• The ball will arrive first on the ground, and it will also be moving faster
just before it reaches h=0. However, they will have equal kinetic
energy because the rotation kinetic energy of the hoop makes up for
this decrease in speed. [This has it all right, but there is no explanation
for the results, let’s look into that]
Parallel-Axis Theorem
Ih = Icom +
2
Mh
h
Rotation Axis
throurgh COM
Parallel
Axis a
distance
h away.
Chapter 10 problems
Chapter 10 problems
•
•
•
•
The mass of the Earth is about 6x1024 kg and its radius is
about 6x106 meters. Suppose you build a runway along the
equator, line-up a million F-16's, bolt them down, and have
them all fire their engines (eastward) simultaneously for 1/2
hour. Estimate the effect that would have on the rotational
speed of the Earth. Assume the thrust of each plane's
engine (these fighters have only one) is 30,000 lbs =
133,000 Newtons. [31 no answers; 11 confused]
very few people knew even how to handle the angular
version of Newton II (t = I a )
Ft = mra (??)1.33x10^11 x 6x10^24 x 6x10^6=a a=
4.788x10^42 [where did this eqn. come from? Dealing with
angular motion, but mass and accel (not I and a) are the
focus of this answer??]
The net torque on the earth from the jets is 6x10^6m x
133000N x 1000000 = 7.98x10^17 Nm. Using t net = inertia
x angular acceleration, I found the acc. to be 9.236x10^-21
rad/s/s. After 30 minutes, the speed would only increase by
1.66x10^-17 rad/s. [4 were like this; this missed the basic
point, but did handle the t = I a correctly]
•
The mass of the Earth is about 6x1024 kg and its radius is about 6x106 meters.
Suppose you build a runway along the equator, line-up a million F-16's, bolt them
down, and have them all fire their engines (eastward) simultaneously for 1/2 hour.
Estimate the effect that would have on the rotational speed of the Earth. Assume
the thrust of each plane's engine (these fighters have only one) is 30,000 lbs =
133,000 Newtons. [31 no answers; 11 confused]
•
very few people knew even how to handle the angular version of Newton
II (t = I a )
Ft = mra (??)1.33x10^11 x 6x10^24 x 6x10^6=a a= 4.788x10^42 [where
did this eqn. come from? Dealing with angular motion, but mass and
accel (not I and a) are the focus of this answer??]
The net torque on the earth from the jets is 6x10^6m x 133000N x
1000000 = 7.98x10^17 Nm. Using t net = inertia x angular acceleration, I
found the acc. to be 9.236x10^-21 rad/s/s. After 30 minutes, the speed
would only increase by 1.66x10^-17 rad/s. [4 were like this; this missed
the basic point, but did handle the t = I a correctly]
•
•
•
HOWEVER: The key is really that the torque from
the F-16’s is INTERNAL to the Earth/atmosphere
system, so there is NO impact on the angular
velocity
Rotational Inertia for Selected objects and rotation axes: HR&W Table 10-2
Chapter 11
Why is it “easier” to do
a dive in the tuck position
than the pike position?
(Note: the diagram shows a
dive in the pike position, not
the tuck as the book suggests).
Chapter 11 problems
b) The KE increases; where did this energy come from?
Rotational Inertia for Selected objects and rotation axes: HR&W Table 10-2
Chapter 11 problems
Chapter 11 problems
A
P
•
B
P
C
P
Rank the three cases in order of the magnitude of the torque gravity exerts
about the point where the post is attached to the ground (P). The cable
supporting the ball is the same distance from the point P in each case, and
the height at which the cable is attached to the supporting structure is also
the same in all three cases.
• B>A>C (6); A>B>C (5) B>C>A (3); C>B=A (3);
• The rankings in order are C then B then A. This is
because C forms a right angle so therefore has the
greatest torque. Secondly, B has a greater distance
between the ball and the hanger therefore it has a
greater moment of inertia. [Be careful of what angle you
use]
• The order is B›A›C. In each case the torque exerts
greater influence for a more horizantally directed
rotation. In the case of b and a more torque is directed
along b's rotational axis because there is a greater
tendency to rotate outward [What direction?]
A
P
•
B
P
C
P
Rank the three cases in order of the magnitude of the torque gravity exerts
about the point where the post is attached to the ground (P). The cable
supporting the ball is the same distance from the point P in each case, and
the height at which the cable is attached to the supporting structure is also
the same in all three cases.
• C>B>A (6); B>A>C (4); B>C>A (3); A>B>C (3)
• In fact: A=B=C!! (neglecting mass of cable and
rods).
– R is the vector from the axis to the point the force is
applied; This is the same for all three, and the
direction (and magnitude) of the gravitational force is
the same in all.
Chapter 12 problems
Chapter 12 problems
Chapter 12 problems