Transcript المحاضرة الثالثة Circular Motion

‫تابع المحاضرة الثالثة‬
Circular Motion
There are two types of circular motion:1- Uniform circular motion
2- Non uniform circular motion
1- Uniform circular motion
Quiz 1
1-Write the dimension of the following:-
Pressure - density – Work
2-check the following equation
a- V=squar root(2*R*g) where
R :radius of the planet
g :the gravitational acceleration at its surface
b- A car with a mass 2500kg moving with
velocity 2ookm/hr ,a break force applied on it
to stop after 5 sec find
The acceleration
The distance traveled before it stopa
Circular motion
If an object or car moving in a circular path with
constant speed V such motion is called uniform
circular motion , and occurs in many situations.
The acceleration a depend on the change in the velocity
vector
a =dv/dt
Because velocity is a vector quantity, there are two
ways in which an acceleration can occur:-
C.M
1- Changing in magnitude of velocity
2- changing in the direction of velocity
For an object moving with constant speed in a circular
motion a change in direction of velocity occurs.
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The velocity vector is always tangent to the path of
the object and perpendicular to the radius r of the
circular path.
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The acceleration vector in uniform circular motion
is always perpendicular to the path and points
toward the center of the circle and called
centripetal acceleration ac
C.M.
ac=v**2/r where
V= velocity
r= radius of circle
 If the acceleration ac is not perpendicular to the
path, there would be a component parallel to the
path and also the velocity and lead to a change in the
speed of the particle and this is inconsist with
uniform circular motion.
 To derive the equation of acceleration of circular
path consider the following diagram
C.M
V
C.M
The particle at position A at time t1 and its
velocity is Vi
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At position B the velocity Vf at time t2
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The average acceleration ac is
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ac=Vf-Vi / t2-t1
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The above two triangle are similar and we can
write a relationship between the length of the
triangles as follow
Δv /v=Δr/r
Where v= vi= vf and r = ri=rf
a =ΔV/Δt = v/r* Δr/Δt
ac=v/r*v=v**2/r
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C.M
linear velocity
angular velocity
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V=distance/ time
t
/ϴ=ω
ϴ =ωt
V=distance /time in m/sec
ω=ϴ/t
ϴ=ωt
after one complete cycle the time is the periodic
time T and
V=2πr/T
(1) and ω=2π/T (2)
From (1) , (2)
C.M
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V= ωr
The angle ϴ swept out in a time t is
ϴ=(2π/T)*t = ωt
(3)
Angular acceleration ac=dωr/dt
ac= ωdr/dt= ωv but v= ωr
ac= ω2r
C.M.
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Non uniform circular motion
If the motion of particle along a smooth curved path
where the velocity is changes in magnitude and
direction .
As the particle moves along curved path the direction
of the acceleration changes from point to point.
The acceleration a can be resolved into two
component :1- ar along the radius r
2- at perpendicular to r
C.M
a**2=(at )**2+(ar )**2
Where
1- at the tangential acceleration component causes
the change in the speed of the particle this
component is parallel to the instantaneous velocity
and given by
at = dv/dt
2- ar is the radial acceleration component arises from the
change in direction of the velocity vector and is given by :
C.M
ar= -ac =-v2/r
in uniform circular motion ,where v is constant ,at
=0 and the acceleration is always completely
radial.
If the direction of v does not change, then there is
no radial acceleration ar=0 and the motion is one
dimensional (ar=0 but at≠0)
Simple Harmonic Motion
Simple harmonic motion is a back and forth
motion caused by a force directly proportional
to the displacement
F αX
F = -KX where
F : restoring force
K: spring constant
X: displacement on the spring
∑F=0
S.H.M
When an object vibrate or
oscillates back and forth over
the same bath and taking the
same time called periodic time
The simplest form of periodic
motion is represented by an
object oscillating on the end
of uniform spring
S.H.M
The object of mass m slides on the
horizontal surface.The position of the
mass m if no force exerted on it is called
equilibrium point x=0
If the mass is moved either to the left, which
compressed the spring or to the right
,which stretches it, the spring exerts a
force on the mass that acts in the direction
of restoring the mass to equilibrium and
called restoring force
S.H.M
S.H.M
The magnitude of the restoring
force F is directly proportional to
the displacement x the spring has
been stretched or compressed
F=-kx
F∞x
The minus sign indicate that the restoring
force F is always in the direction opposite
to the displacement X
S.H.M
The greater value of K , the greater
the force F needed to stretched
or compressed the spring.
The force F is not constant but
varies with the position X, also
the acceleration is not constant
To discuss vibration motion we need
to define the following terms:-
S.H.M
1- Displacement x
The distance of the mass m from the equilibrium point
at any moment
2- Amplitude A
The maximum displacement from the equilibrium
point
3-Frequency f
The number of complete cycle per second
Time period T
The time required to complete one cycle
F=1/T
S.H.M
Energy in the simple harmonic oscillator
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4- E total mechanical energy
is the sum of kinitic and potential
energy
1- Potential energy (P.E)= 1/2 K X2
2- Kinetic energy (K.E) =1/2 m V2
E=K.E+P.E=1/2(mv2+kx2)
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S.H.M
S.H.M
1- at X=0 total mechanical energy is kinetic energy , v
= vmax and P.E=0
E=K.E+P.E=I/2m(Vmax)2+0
(1)
at x=A or at x=-A
The total mechanical energy Eis potential energy and
K.E=0 and V=0 equal Vmin
E=P.E+K.E =1/2(KA2)+0
(2)
From (1),(2)
Vmax=A
S.H.M
To calculate the velocity of mass at any point
between equilibrium and where X=0 and amplitude
where X=A OR X= -A
E=K.E+P.E =1/2(mv2+KX2)=1/2KA2
1/2Mv2=1/2KA2-1/2KX2
mv2=k(A2-X2)
V2=K/m( A2-X2)
V2=(K/m)A2 (1-X2/A2)
Example
 Horizontal spring if a force of 6N is applied
a displacement of 0.03m is produced, when
0.5 Kg is attached to the spring and
stretched for 0.02m and release to oscillate
find: Force constant
2)Angular Velocity
 3) Frequency
4)Max. Velocity
 7)Velocity at displacement 0.01m
 8)Acceleration at displacement 0.01m
 9)Total K.E
10)Total P.E
 11)Total mechanical energy Etotal
Sol.
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Sol.
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1) F =-KX 6 =-K .0.03 K=6/0.03 = 200
2) ω =
=
=20
3) F = 𝝎 /
=3.184 Hz
4) Vmax =A
=
A(
) = 0.02
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5) T=1/ f =1 .3.184=0.314 sec
6) amax = Fmax / m K.A/m =200x0.02 /0.5 =8 m/sec
7) V=Vmax
=0.4
/0.02^2 =
Sol.
8) a =F /m =K.X /m =200 x 9.91 /0.5
=4 m/s2
9) K.E max =0.5 .K.A2 =0.5 .200.0.02^2
= 0.04 J
10) P.Emax =(1/2 )K A^2 = 0.5
.200.0.02^2 =0.04 J
11) Etotal = K.E+ P.E=0.08 J
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Simple pendulum
It consist of a small object suspended
from the end of a light weight cord.
The motion of a simple pendulum swinging
back and forth with negligible friction
If the restoring force F is proportional to the displacement
x ,the motion will be simple harmonic motion.
The restoring force is the net force on the
end of the bob (mass at the end of the
pendulum) and equal to the component of
the weight mg tangent to the arc.
S.H.M
F=-mgsin ϴ
For small angles sin equal angle
ϴ=sinϴ
F=-kx and F=- mgϴ , ϴ=X/L
K=mg/L or k/m=g/L
S.H.M
The time constant T derived as follows:F=-mg x/L =ω2r m
ω2=-g/L
ω =2πf
4π2f2=g/L
f2 =(1/4) π2 (g/L)
f=(1/2π)
The time period T = 1/f
T=2π
Exercise
Car with mass 1000Kg when person of
weight 980 N climb slowly into the car
sinks 2.8 cm and vibrates up and down
find:
1- Constant of the force K
2- Angular velocity 𝝎
3- Frequency f
4- Periodic time T
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