تابع المحاضرة الثالثة circular motion there are two types of circular motion:- 1-...

الثالثة المحاضرة تابعCircular 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 – Work2-check the following equationa- V=squar root(2*R*g) whereR :radius of the planetg :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.M1- Changing in magnitude of velocity2- changing in the direction of velocityFor an object moving with constant speed in

a circular motion a change in direction of velocity occurs.

The velocity vector is always tangent to the path of the object and perpendicular to the radius r of the circular path.

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 At position B the velocity Vf at time t2 The average acceleration ac is

ac=Vf-Vi / t2-t1 The above two triangle are similar and

we can write a relationship between the length of the triangles as follow

Δv /v=Δr/rWhere v= vi= vf and r = ri=rf a =ΔV/Δt = v/r* Δr/Δt ac=v/r*v=v**2/r

C.M

linear velocity V=distance/ time angular velocity 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

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.

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 α X 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 xThe distance of the mass m from the

equilibrium point at any moment2- Amplitude AThe maximum displacement from the

equilibrium point3-Frequency fThe number of complete cycle per second Time period TThe time required to complete one cycle F=1/T

S.H.M

Energy in the simple harmonic oscillator 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)

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=-AThe 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. Sol.

1) F =-KX 6 =-K .0.03 K=6/0.03 = 200 2) ω = = =20 3) F = / =3.184 Hz𝝎

4) Vmax =A = A ( ) = 0.02

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 

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 ϴFor small angles sin equal angle ϴ=sinϴ

F=-kx and F=- mgϴ , ϴ=X/LK=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/Lω =2πf 4π2f2=g/Lf2 =(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