الثالثة المحاضرةCircular 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 equation

a- 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 findThe 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 velocity

2 -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.Mar= -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)