Download Torque

Chapter 6
Motion in Two
Dimensions
Merrill Physics: Principles and
Problems
A. Motion in two dimensions
1. Will use the kinematic
equations:
d = vt
d = vot + 1/2 at 2
vf2 = vo2 + 2ad
vf = vo + at
d = {(vf + vo ) / 2}(t)
2. Will use Newton’s Laws
1st: An object in motion remains in
motion unless acted on by a net
force
2nd :
F=ma
where F is net force
3. And coefficients of friction:
 f f / F N
Projectile Motion
Objectives:
 Understand that the vertical and
horizontal components of a projectile
are independent
 Determine the range, maximum
height and flight time of projectiles
*As you throw a ball, or any type of object,
what shape does it make in the air?
Definitions used for Projectile Motion
(Ballistics)
a. Projectile - object that only
has gravity acting on it
b. Trajectory - path a projectile
follows
c. Shape of the trajectory depends
on your point of view and the
object.
(1) Top looking down on object’s
path
ending point
starting point
What force accelerates the object in the x-direction?
(None once you let go or the bullet is fired)
*this assumes no air resistance for simplicity
So velocity is constant in the x-direction
(2) Side view of object’s motion
start
end
What force accelerates in the y-direction?
Gravity!
So VY is a changing value.
4. Three assumptions used in
analyzing motion
Assumptions:
(1) acceleration due to gravity is
9.8 m/s2 down
(2) air resistance is negligible
(3) rotation of the earth does not
affect motion
This is not rocket science! ….
Otherwise we would include these!
Independence of Motion in Two
Dimensions

What forces act on an object after
it’s initial thrust if you ignore air
resistance?
Projectiles launched at an angle
Objectives:

Determine the range, maximum
height and flight time of
projectiles launched at an angle
Projectiles Launched Horizontally
Example: A stone is thrown horizontally off a
44m cliff at velocity of 15 m/s. How long
will it take to hit the bottom? How far will it
go horizontally if the air resistance is
negligible?
Given:
vx = 15 m/s
voy= 0m/s
ay = -9.8 m/s2
dx = vxt = 15m(t)
dy = voyt + ½at2
44m = ½(9.8m/s2)t2
* Time is the same in the vertical and horizontal
directions!
t2 = 2(44m)/9.8m/s2
t = 3s and 15m/s(3s) = 45m
Projectiles Launched at an Angle
b.
(1) velocity in x-direction never changes
(2) velocity in y-direction changes
vy-f = vy-o + g t
vy-f2 = vy-o2 + 2 gdy
(3) vR2 = vx2 + vy2
Projectile Motion Problem Solving
a. Projectile is an object that only has
gravity acting on it
b. Break given initial velocity into x
and y components. v at an angle q
(1)horizontal motion: vx = v cosq
(2)vertical motion: vy = v sinq
c. use subscripts to identify x & y
components of velocity, displacement,
acceleration
d. Use independence of vertical and
horizontal motion to solve problems
e. Draw diagrams to aid you in
visualizing what is occurring
f. Write down given information with
signs
g. Note: ax = 0
ay = - 9.8 m/s2
- time is the same for both directions
- vertical motion is symmetrical!
Velocity of the Projectile

The velocity of the projectile at any
point of its motion is the vector sum
of its x and y components at that
point
2
x
v  v v
2
y
and
q  tan
1
vy
vx
• Remember to be careful about the
angle’s quadrant
Example: Object is launched at an angle of
60o at a velocity of 110 m/s. How far will it
go if the air resistance is negligible?
Given:
vx = 110 cos 60 = 55 m/s
voy = 110 sin 60 = 95.3 m/s
vfy = - 95.3 m/s
ay = -9.8 m/s2
* Commonality is with TIME. How long it is in
the air is same for horizontal and vertical
motion. Knowing time, you can find
distance in x
vyf = v0y + ayt
t = 19.4 sec use travel in the vertical
direction to determine how long
object is in flight. Remember, object
is accelerated in the y-direction.
Knowing time then can find out how far
vx = dx / t since uniform motion in xdirect.
dx = (55 m/s) (19.4 sec) = 1070
meters
i. Example Problem:
A punter kicks a football at an
angle of 30o with the horizontal at
an initial speed of 20 m/s. Where
should the punt returner position
himself to catch the ball just
before it strikes the ground?
Givens:
vy = 20 m/s sin 30 = 10 m/s
vx = 20 cos 30 = 17.3 m/s
vy 0 = 10 m/s
v
30
vyf =-10m/s
vyf = vy 0 + at
where a=-9.8m/s2
t = 2.04 sec
dx = vx t =17.3 m/s (2.04s)
= 35.3 m
Projectile Motion Simulation labs


http://phet.colorado.edu/en/simulati
on/projectile-motion
http://phet.colorado.edu/sims/projec
tile-motion/projectilemotion_en.html
Periodic Motion




Explain centripetal acceleration of an
object in circular motion
Apply Newton’s Laws to circular
motion
Define simple harmonic motion
Apply Newton’s Laws to harmonic
motion
B. Periodic Motion
1. Uniform Circular Motion
a. Consider a model airplane on
control strings - the hand pulls on
cable and keeps an inward force
on the plane so it flies in a circle.
Let go of the string and it flies in a
straight line
A net force, acting perpendicular to a
mass moving at constant speed,
changes direction of motion of the
mass.
(1) Force is always 90o to
direction of motion of the mass
(2) Since it is perpendicular, cannot
change the magnitude of
velocity only direction.
(3) Net force is centripetal force,
Fc.
Change in direction is a change in
velocity (vso object is
accelerated
(1)Fc is towards the center, so
acceleration is towards the
center.
(2) Called centripetal acceleration.
Centripetal Acceleration Formulas:
(1)
ac = v2/r
(2)
Fc = mac = mv2/r
Key: Centripetal force is always
caused by some other action friction, pulling, pushing, etc.
We define period of circular motion
(T) to be the time to go around the
circumference one time
For uniform circular motion time to
go around circle is constant
v = d/t (d = the circumference)
where d = 2r & t = T (the period)
therefore v2r/T
and
ac = v2/r = 4r/T2
Fnet = Fc = mac = mv2/r = m(4r/T2)
Example: A person is flying an 80.0g model
airplane in a horizontal path at the end of a
string 10.0m long. If the string is horizontal
and exerts a 2.65N force on the hand, what
is the speed of the plane?
Givens:
m= 0.080kg
r = 10.0 m
V = (Fcr/m)1/2 = 18.2 m/s
Formulas
v = 2r/T
Fc = mac = mv2/r
A 200g object is tied on a cord and
whirled in a horizontal circle of radius
1.20m at 3 rev/sec. The cord is
horizontal. Determine acceleration
and tension in the cord.
Givens:
Formulas
m= 0.200kg
v = 2r/T
r = 1.20m
rev/sec (freq.)
1/ 1/ sec
vm/s
a. ac = v2 / r = 426 m/s2
b. Fc = m ac = 85 N
What is the maximum speed at which a
car can round a curve of 25m radius on a
level road if static between the tires and
road is 0.80
Fc = Ff =  static FN = static m(-g)
Fc = mac = static m(-g) = mv2/r
Solving for v: v = (rstatic-g)1/2
Therefore v = (25mx0.80x9.8m/s2)1/2
v = (196m2/s2)1/2 = 14 m/s
* the maximum velocity only depends on the
radius of the curve and static , not the mass!
11.1 Torque
To make an object turn or
rotate, apply a torque.
Torque
Every time you open a door, turn on a water
faucet, or tighten a nut with a wrench, you
exert a turning force.
Torque is produced by this turning force and
tends to produce rotational acceleration.
Torque is different from force.
• Forces tend to make things accelerate.
• Torques produce rotation.
Torque
A torque produces rotation.
Changing Circular Motion (Torque)


The door is free to rotate about an axis
through the hinge at point O
There are three factors that determine the
effectiveness of the force in opening the
door:
• (1) The magnitude of the force
• (2) The distance of the application of the force
• (3) The angle at which the force is applied
Torque

Magnitude of force

Distance from axis

Direction
Torque
When a perpendicular force is applied, the
lever arm is the distance between the
doorknob and the edge with the hinges.
Torque
When the force is perpendicular, the distance from
the turning axis to the point of contact is called
the lever arm.
If the force is not at right angle to the lever arm,
then only the perpendicular component of the
force will contribute to the torque.
Torque
The same torque can be produced by a
large force with a short lever arm, or a
small force with a long lever arm.
The same force can produce different
amounts of torque.
Greater torques are produced when both
the force and lever arm are large.
Torque


t = Frsinθ
Units – Nm

If θ = 90o
Then sinθ = 1

t = Fr or Fd where d = the lever arm

length

Torque, t, is the tendency of a force to
rotate an object about some axis
t = Frsinq
t
is the torque
F is the force
t symbol is the Greek tau
rsinq is the length of the lever arm acting
perpendicular to the force where q is the angle
between F and r
SI unit is N.m
Torque
Although the magnitudes of the applied forces
are the same in each case, the torques are
different.
Torque
think!
If you cannot exert enough torque to turn a stubborn
bolt, would more torque be produced if you fastened
a length of rope to the wrench handle as shown?
Answer:
No, because the lever arm is the same. To increase
the lever arm, a better idea would be to use a pipe
that extends upward.
Balanced Torques
do the math!
What is the weight of the block hung at the 10-cm
mark?
Balanced Torques
do the math!
The block of unknown weight tends to rotate the
system of blocks and stick counterclockwise, and
the 20-N block tends to rotate the system
clockwise. The system is in balance (static
equilibrium) when the two torques are equal:
counterclockwise torque = clockwise torque
Balanced Torques
Rearrange the equation to solve for the unknown weight:
The lever arm for the unknown weight is 40 cm.
The lever arm for the 20-N block is 30 cm.
The unknown weight is thus 15 N.
Torque problems

Complete Gantry Crane Worksheet
(Due Thurs 1/7/2014)
Simple Harmonic Motion
Objectives


Identify objects in simple harmonic
motion.
Determine variables that affect the
period of a pendulum.
Simple Harmonic or Periodic Motion

Examples – clock pendulum,
vibrating guitar string
Simple Harmonic Motion

Described by two quantities:
• Period
• Amplitude
Harmonic Motion in a Spring
Simple Harmonic Motion
A weight attached to a spring undergoes simple
harmonic motion.
A marking pen attached to the bob traces a sine
curve on a sheet of paper that is moving horizontally
at constant speed.
A sine curve is a pictorial representation of a wave.
A sine curve is a pictorial representation of a SHM.

Motion of a Pendulum
A stone suspended at the end of a string is a simple
pendulum.

Pendulums swing back and forth with such regularity
that they have long been used to control the motion of
clocks.

The time of a back-and-forth swing of the pendulum
is its period.

Galileo discovered that the period of a pendulum
depends only on its length—its mass has no effect.

Motion of a Pendulum
The back-and-forth vibratory motion—called
oscillatory motion—of a swinging pendulum is
also simple harmonic motion (SHM).

T  2
l
g
The period of the pendulum depends only on the
length of a pendulum and the acceleration of gravity.
Motion of a Pendulum
Two pendulums of the same length have
the same period regardless of mass.
Example Problem
How long must a pendulum be on the
Moon, where g=1.6 m/s2, to have a
period of 2.0 s?
T = 2.0s = 2(l/1.6m/s2)1/2
(2.0s/2)2 = l/1.6m/s2
l = 1.6m/s2(4s2/42) = 0.16 m or 16cm
resolved into components
The weight w can be ______________________________
:
q q
w||
w┴
w||
w┴
w
equilibrium
This component
perpendicular
is ______________
to the motion and
opposite
is ____________
in
direction to the
tension
____________
w
parallel to the
w|| is the part of w that is ______________________motion.
increases
w|| _______________
as q increases
equilibrium
w|| pulls the bob back towards __________________
restoring
w|| is called a ___________________
force.
Motion of a Pendulum
A long pendulum has a longer period than a shorter
pendulum.

It swings back and forth more slowly—less
frequently—than a short pendulum.

Just as a long pendulum has a greater period, a
person with long legs tends to have a slower stride
than a person with short legs.

Giraffes and horses run with a slower gait than do
short-legged animals such as hamsters and mice.
