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المحاضرة الثالثة Circular Motion

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

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المحاضرة الثالثة Circular Motion

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  1. المحاضرة الثالثةCircular Motion There are two types of circular motion:- 1- Uniform circular motion 2- Non uniform circular motion 1- Uniform circular motion

  2. 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

  3. 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:-

  4. 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. • 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

  5. 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

  6. C.M V

  7. 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/r Where v= vi= vf and r = ri=rf a =ΔV/Δt = v/r* Δr/Δt ac=v/r*v=v**2/r

  8. C.M linear velocityV=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)

  9. 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

  10. C.M. • Non uniform circular motion • Ifthe 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

  11. 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 :

  12. 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)

  13. 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

  14. 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

  15. 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

  16. S.H.M

  17. 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

  18. 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:-

  19. 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

  20. 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)

  21. S.H.M

  22. 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

  23. 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)

  24. 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

  25. Sol. • Sol. • 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 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 =

  26. 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

  27. 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.

  28. 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

  29. 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

  30. 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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