Circular motion,Vertical circular motion and Radius of curvature

If Any object moves such that it covers an angle q at the fixed point (center of the circle) and its gap from the fixed point remains constant (radius of the circle)

 

Here R is constants

 

     

Now finding linear speed of

(i) hour hand (30 cm)

(ii) Minute hand (90 cm)

(iii) Second hand (60 cm)

Solution:- (i) time period of hour hand T = 12 hr

               = 12 x 3600 sec

                                                                v = r w

 

(ii) Minute hand:- time period = T = 60 min = 60 x 60 sec

 

(iii) Second hand:-

Time period= T = 60 sec

v = r w

Types of circular motion: (on the basis of speed)

(1) Uniform circular motion: (U C M)

Only direction of velocity is changing, magnitude remains unchanged.

 

Now acceleration due to change in direction of velocity

Using law of parallelogram

 

 

 

= 2v² - 2v² cosq                                 

= 2v²  (1- cosq)

 

 

Now

 

 

sinq » q if  q <<<<

 

 

a = v w,

 

 

 

Direction of this acceleration is towards center

   

Centripetal acceleration or Radial acceleration

(2) Non uniform circular motion (Non U C M)

Here in this case direction as well as magnitude both are changing continuously therefore here two different named acceleration will be as,

(i) a_c = a_R (centripetal or radial acceleration)

(ii) a_t (Tangential acceleration)

Now,

Tangential acceleration. 

“Rate of change of magnitude of velocity”

v = R w

 

 

 

So a_t ^ a_c

 

Net acceleration for non-uniform circular motion.

Example:- A particle moving in a circular path of radius 2 meter and its velocity varies as  v = 10t^2. Then net acceleration of the particle at t = 2 sec.

Solution: Given v = 10t²

 

= 20 x 2

= 40 m/s²

 

 

Centripetal force:- the force require to move in a circular path for any object with respect to inertial frame is called centripetal force.

 

Note:- Centrifugal force will also have same magnitude but direction opposite to centripetal force.

Direction of this force is towards the center.

HORIZONTAL CIRCULAR MOTION

(1) Only banking

Here N sinq is C.P. provider

 

But N cosq = mg

 

     

            Safe speed

(2) Only friction 

Here now m N is C.P. provider

 

N = mg

 

 

 

Safe speed

Friction and Banking both:- To prevent inward sliding

 cosq + m N sinq = mg

Divide

          

TO PREVENT OUTWARD SLIDING

N cosq - m N sinq = mg

          

 

v₁ < V_safe < v₂

CONICAL PENDULUM

T sinq = mrw²

T cosq = mg

 

 

Time period of conical pendulum