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. 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

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

C.M linear velocity V=distance/ time angular velocity t / =ω =ωt V=distance /time in m/sec ω=/t =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 swept out in a time t is =(2π/T)*t = ωt (3) Angular acceleration a c =dωr/dt a c = ωdr/dt= ωv but v= ωr a c = ω 2 r

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=(a t )**2 +(a r )**2 Where 1- a t the tangential acceleration component causes the change in the speed of the particle this component is parallel to the instantaneous velocity and given by a t = dv/dt 2- a r is the radial acceleration component arises from the change in direction of the velocity vector and is given by :

C.M a r = -a c =-v 2 /r in uniform circular motion,where v is constant,a t =0 and the acceleration is always completely radial. If the direction of v does not change, then there is no radial acceleration a r =0 and the motion is one dimensional (a r =0 but a t ≠0)