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Motion

Distance(d)- simply pertains to length (scalar) e.g.
5m, 5km

Displacement (d)- change in position specified by
magnitude + direction (vector) e.g. 5m, north , 5m, 20˚ S
of E.
Speed (v) – distance per unit time (scalar)
 Velocity (v) – displacement per unit time (vector)
-ve velocity is in the opposite direction

v= speed/velocity
𝑣=
𝑑
𝑡
d= distance/displacement
t= time
•
Acceleration – time rate of change of
velocity
𝑣𝑓 − 𝑣𝑖
𝑎=
𝑡
if
Where:
∆𝑣 = 𝑣𝑓 − 𝑣𝑖
a= acceleration
t= time
vi= Initial velocity
vf= final velocity
Uniformly Accelerated Motion: the motion in
a straight line in which the rate changes
uniformly
Equations
1. 𝑑 =
1
2
𝑣𝑖 + 𝑣𝑓 𝑡
2. 𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
3. 𝑑 = 𝑣_𝑖𝑡 +
1
𝑎𝑡 2
2
4. 𝑣𝑓 2 = 𝑣𝑖 2 + 2𝑎𝑑
NEWTON’S LAWS OF MOTION
First law: law of Inertia
a body at rest remains at rest and a body in
motion continues to move in straight line at
constant speed, unless an external unbalanced
force acts on it.
A=B
No movement
A
Unbalance
force
B
A=B
rotate
Second law: Law of Acceleration
an external unbalance force acting on an object
produces an acceleration in the direction of the
net force, an acceleration that is directly
proportional to the unbalanced force and
inversely proportional to the mass of the body
Where:
F=force
m=mass
a= acceleration
Where:
w = weight
m = mass
g = gravity
Third law: Interaction
for every force that a first body exerts upon
second body, there is a force equal in magnitude
but opposite in direction that a second body
exerts upon the first body.
A 100N box is sliding down a frictionless plane
inclined at an angle of 30° from the horizontal.
Find the acceleration of the box.
Note: N is always ┴ to the plane
N
FBD- free body
diagram- shows
all the forces acting
on the object
F
W
Solution:
UNIFORM CIRCULAR MOTION
- motion of an object in a curved or circular
motion
10 rev. in 5s
Frequency, F = the no. of revolution per unit time
Period, T = the time required for one complete
revolution.
Where:
v= linear velocity
r= radius
Radian- angle subtended by the arc of a circle whose length is equal to the
radius of the same circle.
x
r
Where: ω = angular velocity
= angle turned through
t = time elapse
Where: v= linear velocity
r= radius
ω =angular velocity
Using linear velocity
if
Centripetal force
Using angular velocity
SAMPLE PROBLEM
If the radius of the circular path of the stone is
0.5m and its period is 0.5s. What is its constant
speed?
Given: r= 0.5m
T= 0.5s
Required : v=?
Solution:

What is the angular velocity of a stone which
makes 10 rev in 5 seconds? The radius of the
circular path is 0.5m
Given: no. of revolution = 10
T= 5s
r=0.5m
Required: ω=?
Solution:

A mass of 0.5kg is whirled in a horizontal circle of
radius 2m. If it makes 5 rev in 5s
Find: a. speed
b. acceleration
c. centripetal force
Given: m=0.5 kg
r=2m
t=5s
rev=5
T=1s

Solution:
End of presentation