Transcript Free Fall

Kinematics
Motion Equations
Equations of Motion
Constant Acceleration
Constant Acceleration Problem Solving
Centripetal and Tangential Acceleration
Free-Fall Motion
1
Kinematics
Motion Equations
Motion can be determined by using a few simple equations.
The relationships between velocity and position are…
r
v
t
rf  r0  v t
The relationships between acceleration and velocity are…
v
a
t
2
v f  v0  a t
Kinematics
Motion Equations
If the acceleration is constant, we see that these become.
v  at
1
r  v0 t  a t 2
2
Combining these, we find another equation.

 
v f  v0  at
v f v f  v0 v0  2v0 at  a at 2
1


v f v f  v0 v0  2a  v0 t  a t 2   v0 v0  2a r
2


v f v f  v0 v0  2a r
3
Kinematics
Motion Equations
We can use these three equations to solve for any motion involving
constant acceleration.
v  at
This equation relates velocity and time.
1
r  v0 t  a t 2 This equation relates position and time.
2
equation relates position and
v f v f  v0 v0  2a r This
velocity.
4
Kinematics
Motion Equations
If we are only dealing with one vector component, then the equations
become simple.
Let’s just look at the x-component
v  at
vx  ax t
1
r  v0 t  a t 2
2
v f v f  v0 v0  2a r
1
x  vx 0 t  ax t 2
2
vxf 2  vx 02  2ax x
The y-component and z-component equations are similar.
5
Kinematics
Motion Equations
Now let’s see how we use them.
Example: Two race cars are moving on a racetrack. The lead car
is ahead by 10 m. Both cars are currently moving at 100 km/hr. If the
second car accelerates at 10 m/s2, how long will it take to reach the
lead car?
r1
v1 f  100 km/hr
a1  0
r01  10m
v10  100 km/hr
v20  100 km/hr
a1  0
a2  10
v2 f
2
m
a2  10 2
s
6
r2
click the icon to
open the worksheet
m
s2
Kinematics
Motion Equations
Now let’s see how we use them.
r1
r01  10m
v10  100 km/hr
v20  100 km/hr
a1  0
a2  10
2
r2
7
v1x  a1x t1
v2 x  a2 x t2
1
x1  v1x 0 t1  a1x t12
2
v1xf 2  v1x 02  2a1x x1
1
x2  v2 x 0 t2  a2 x t2 2
2
v2 xf 2  v2 x 02  2a2 x x2
m
s2
Kinematics
Motion Equations
Now let’s see how we use them.
r1
r01  10m
v10  100 km/hr
v20  100 km/hr
a1  0
a2  10
m
s2
2
r2
km 

0
vv11xfx  a1100
x t1hr 


8
km 1

2

t
xx11  v1x100

t

a

t
0
1  1 1x
1
hr  2

2
22 
2 km 
v1v1xfxf 
x
 v1x100
0  2a
1x 1
hr 

km  
m

v
100 t    10 2  t2
2 xfv
2 x  a2 x 
hr 2 
s 

km  1 1 

2m 
2
xx


100

t


10

t

v

t

a

t
22 
2
 2

2 
 2 x 0 hr 2 2 22 x 2 s 
2
km
m



2
v2vxf22xf2 
100

2

10
v2 x 0 2a2 xx2 2  x2
hr 
s 


Kinematics
Motion Equations
Now let’s see how we use them.
r1
t1  t2  t
x2  x1  x01
r01  10m
v10  100 km/hr
v20  100 km/hr
a1  0
a2  10
m
s2
2
r2
v1xf
9
km 

  100
0
hr 

km
km 
m
m

vv22xfxf 100
100  10
10 22 tt2
hr
hr  
ss 

km 

km
km
km
1
m  22

x2 x1 10
m


100

t

100   t1
xx2 2100
100 t2t   10 2  t 2

hr


hr 
hr
hr 
2
s 


2
2
km
km
m





2
v1xf 2   100
v


100

2

10
x2
2 xf




2 
hr 
hr 
s 



Kinematics
Motion Equations
Now let’s see how we use them.
r1
t1  t2  t
x2  x1  x01
r01  10m
v10  100 km/hr
v20  100 km/hr
a1  0
a2  10
2
r2
v1xf
10
km  
m

  100
   10 2  t
hr  
s 

km 
1
m 2

x1   100
 t   10 2  t
hr 
2
s 

2
km
m



v1xf 2   100

2

10
x1


2 
hr 
s 


km 

v2 xf   100
0
hr 

km 

x1   10 m    100
 t
hr 

2
km


v2 xf 2   100

hr 

m
s2
Kinematics
Motion Equations
Now let’s see how we use them.
r1
t1  t2  t
x2  x1  x01
r01  10m
v10  100 km/hr
v20  100 km/hr
a1  0
a2  10
2
r2
11
km 
1
m  2 x   10 m    100 km  t

x1   100
1


 t   10 2  t
hr


hr 
2
s 

km
km
1
m
 10 m    100  t   100  t   10 2  t 2
hr 
hr 
2
s 


m
s2
Kinematics
Motion Equations
Now let’s see how we use them.
r1
t1  t2  t
x2  x1  x01
r01  10m
v10  100 km/hr
v20  100 km/hr
a1  0
a2  10
2
r2
12
10 m 

1
m 2
2
 10 m    10 2  t kmt  2 mkm  1t  1.44ms  2
 102 m
   s100

 t  100
 t   10 2  t
10

hr 
2
s 


s 2hr 
m
s2
Kinematics
Motion Equations
What Happened to Centrifugal Force?
There is no such thing as centrifugal force.
So where did it come from?
A mistaken assumption is made that the forces on particles
moving in a circle with constant speed have no forces acting
on them.
Why is this false?
Acceleration comes from changes in velocity (direction, not
just speed).
Circular motion requires acceleration and thus requires force.
13
Kinematics
Motion Equations
What Happened to Centrifugal Force?
The myth starts from the mistaken idea that there is no
acceleration and therefore the total force is zero.

T

Fcentifugal
 

T  Fcentifugal  ma  0
14
Kinematics
Motion Equations
What Happened to Centrifugal Force?
The fact is that velocity is changing and the acceleration is the
centripetal acceleration. The force is NOT zero!!!!

T


T  macentripetal
15
Kinematics
Motion Equations
Centripetal vs. Tangential Acceleration
Centripetal acceleration causes a particle to change its
direction.
v
2
v
ac 
r
r
It points toward
the center of the
circle
16
Kinematics
Motion Equations
Free Fall
Any particle, subject only to the force of gravity is in free-fall.
If an object is in free-fall and we define the positive y-axis as
upward, then its acceleration is always given by

a   g ĵ
where g is the acceleration due to gravity and has a value of
9.81 m/s2 near the surface of the earth.
Note that the acceleration parallel to the earth’s surface is zero.
17
Kinematics
Motion Equations
Free Fall
Particle’s in free fall are subject only to the force of gravity.
Every particle in free-fall has an acceleration of 9.81 m/s2 downward.
The motion
diagram for
any object in
free-fall that
starts from
rest is the
same.
18
Kinematics
Motion Equations
Free Fall
For particle’s that do not start at rest…
1.The vertical acceleration is 9.81 m/s2 downward.
2.The horizontal acceleration is zero. (The horizontal velocity is constant.)
In other words, objects move in a very predictable way.
But then, you already know this.
19
Kinematics
Motion Equations
Free Fall
The motion of a baseball hit at an angle, undergoing free fall is a parabola.
20
Kinematics
Motion Equations
Free Fall
The motion of a rocks thrown from a cliff at different horizontal speeds has
some similarities.
21
Kinematics
Motion Equations
Equations
In free fall, the vertical position, velocity and acceleration are
related by the equations
1 2
y  v0 y t  gt
2
v y   gt
the horizontal position, velocity and acceleration are related by
the equation
x  v0 xt
vx  0
Note that we can write the components of the initial velocity as
22
v0 x  v0 cos  0
v0 y  v0 sin  0