Ch10
NAME: _
HOMEWORK CHAPTER 10:
ROTATION OF A RIGID BODY ABOUT A FIXED AXIS
NOTE:
1.
During a certain period of time, the angular position of a swinging door is described
by:
= 5.00 + 10.0t + 2.00t2, where is in radians and t is in seconds. Determine
Conservative Forces:
As you saw when lifting a book, the work that you do "against gravity" in lifting is stored (somewhere.
Physicists say that it is stored "in the gravitational field" or stored "in the Earth/book system".) and is
available for kinetic
Chapter 10 and 11 Lecture Problems
*Problem 10.1.4: Two basketballs sit sidebyside and are touching. Calculate the
magnitude of the gravitational force the balls exert on each other. The mass of each
basketball is 0.612 kg and its radius is 11.9 cm.
_ N
Michel Georges
Physics
10/20/13
1) watch the youtube video. I give answers to some of the questions. Don't skip that part. COPY AND PASTE
the questions in a doc.
2) plot (scatter lines only + smooth line ) xposition versus time. (title + labels). Include
Chapter 7
Momentum
7.1
What is Momentum?
Momentum is a physical quantity which is closely related to forces. We will learn about this
connection a little later. Remarkably momentum is a conserved quantity. This makes momentum
extremely useful in solving a
include in the nal answer the direction of the objects motion
Worked Example 32
Calculating Momentum 1
Question: A ball of mass 3kg moves at 2m.s1 to the right. Calculate the balls
momentum.
Answer:
Step 1 : Decide what information has been supplied
The
6.5
Important Equations and Quantities
Quantity
Displacement
Velocity
Distance
Speed
Acceleration
Units
Symbol Unit
s
,
u v
s
v
a

Base S.I. Units
m + direction
m.s1 + direction
m
m.s1
m.s1 + direction
Table 6.1: Units used in Rectilinear Motion
125
v1
s1
u2
a1 , a 2
balcony
u1
s2
v2
ground
where the subscript 1 refers to the upward part of the balls motion and the subscript
2 refers to the downward part of the balls motion.
Step 2 : Decide how to tackle the problem
First the ball goes upwards with g
We can nd s1 by using equation 6.4
2
v1
=
s1
u2 + 2a1 s1
1
2
v1 u 2
1
2a
(0m/s)2 (10m/s)2
2(10m/s2 )
5m
=
=
=
Step 6 : For Stage 2, decide what information is supplied
For Stage 2 we have the following quantities:
u2
=
0m/s
v2
a2
=
=
t2
s2
=
=
?
10m/s2
?
Step 6 : Do the calculation
v
=
=
m
m
+ (8 2 )(4s)
s
s
m
32
s
0
Step 7 : Time at half the distance: Find an equation to relate the unknown and known
quantities
Here we have the quantities s, u and a so we do this in 2 parts, rst using equation
6.4 to calc
Step 4 : Do the calculation
Substituting in the values of the known quantities this becomes
a
=
=
=
2(725m 100 m 10s)
s
102 s2
2(275m)
100s2
m
5.5 2
s
Step 5 : Quote the nal answer
m
m
The racing car is accelerating at 5.5 s2 , or we could say it is dece
then equation 6.2 becomes
s
=
=
=
u + u + at
t
2
2ut + at2
2
1
ut + at2
2
Equation 6.4
This equation is just derived by eliminating the time variable in the above equation. From
Equation 6.1 we know
vu
t=
a
Substituting this into Equation 6.3 gives
s
2as
where v is the change in velocity, i.e. v = v u. Thus we have
a
=
v
=
vu
t
u + at
velocity (m/s)
Equation 6.2
In the previous section we saw that displacement can be calculated from the area between a
velocitytime graph and the timeaxis. For uniformly a
Step 2 : The velocity during the last 3 seconds
For the last 3 seconds we can see that the displacement stays constant, and that the
gradient is zero. Thus = 0m/s
v
Worked Example 28
From an accelerationtime graph to a velocitytime graph
Question: Given
Displacement
Velocity
Acceleration
Figure 6.3: A Relationship Between Displacement, Velocity and Acceleration
6.3.4
Worked Examples
Worked Example 25
Relating displacement, velocity, and accelerationtime graphs
displacement (m)
Question: Given the disp
6.4
Equations of Motion
This section is about solving problems relating to uniformly accelerated motion. Well rst
introduce the variables and the equations, then well show you how to derive them, and after
that well do a couple of examples.
u = starting v
time axis
Area
=
=
=
1
bh
2
1
1s 2m/s
2
1m
Step 3 : Determine the total displacement of the car
Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the last second (from t = 14s to t = 15s) the velocity of the
velocity (m/s)
4
15
5
time (s)
12
2
Answer:
Step 1 : Decide how to tackle the problem
We are asked to calculate the displacement of the car. All we need to remember
here is that the area between the velocitytime graph and the time axis gives us the
disp
acceleration (m/s2 )
constant positive acceleration.
Step 8 : Acclerationtime graph  46 seconds
For the nal 2 seconds the object is traveling with a constant velocity. During this
time the gradient of the velocitytime graph is once again zero, and thu
3. Describe the behaviour of the object
velocity (m/s)
Step 3 : Velocitytime graph  02 seconds
For the rst 2 seconds we can see that the displacement remains constant  so the
object is not moving, thus it has zero velocity during this time. We can rea
b) shows the graph for an object moving at a constant acceleration. In this case the acceleration is positive  remember that it can also be negative.
acceleration (m/s2 )
We can obtain the velocity of a particle at some given time from an acceleration ti
6.3.2
VelocityTime Graphs
Look at the velocitytime graph below:
velocity
B
v
t
A
time
This is the velocitytime graph of a cyclist traveling from A to B at a constant acceleration,
i.e. with steadily increasing velocity. The gradient of this graph is ju
Using the rule of Pythagoras:
s
=
=
2
(30m) + (40m)
2
50m in the direction f rom A to C
Step 4 : Now we can determine the average velocity from the displacement and the
time
v
=
=
50m
10s
m
5 in the direction f rom A to C
s
For this cyclist, his velocity
So, here weve shown that an object traveling at 10m/s for 5s has undergone a displacement
of 50m.
Important: The area between a velocitytime graph and the time
axis gives the displacement of the object.
a)
velocity
velocity
Here are a couple more velocit
displacement (m)
50
A
C
s
t
time (s)10
This graphs shows us how, in 10 seconds time, the cyclist has moved from A to C. We know
y
the gradient (slope) of a graph is dened as the change in y divided by the change in x, i.e x .
In this graph the gradient of
Friction
We all know that Newtons First Law states that an object moving without a force acting on it
will keep moving. Then why does a box sliding on a table stop? The answer is friction. Friction
arises from the interaction between the molecules on the
Worked Example 24
Speed and Velocity
40m
C
A
30m
B
Question: A cyclist moves from A through B to C in 10 seconds. Calculate both
his speed and his velocity.
Answer:
Step 1 : Decide what information has been supplied
The question explicitly gives
the dist