Simplified Class Notes for Circular Motion

Kinetics of Circular Motion
As we had discussed in the kinematics of circular
F4 F2 m
motion that there exist two types of acceleration
aC and aT in the general. aC exists because of the
change in the direction only of the linear velocity F1 aR
vector and aT exists because of the change in mag- aT
O F3 O
nitude only of the linear velocity aC always points
towards the center of curvature and a T is directed
along the tangent depending on the whether the
magnitude of linear velocity increases or decreases.
Since ac points towards the center of curvature, there has to be a net positive unbalanced
external force pointing towards the center which gives the body acceleration a c. We call this force as the
centripetal or radial or normal force FC.
For the existence of aT there has to be a net positive unbalanced external force which points
in the direction of aT and we call this force as the tangential force FT.
While discussing the F.B.D. at any position of the body, which is in circular motion, one of the
axis is taken along the radial line and the other along the tangent line.

From the fig. above
2
F1  F2  ma c mv
 R Here (F1 - F2) is called the centripetal force.
F3 - F4 = maT Here (F3 - F4) is called the tangential force.

In the given cases below, we observe centripetal force to be acting under different names.




a) If a stone is tied to a string and whirled in circle then the tension of the string becomes centripetal force.
b) If an electron is revolving around a nucleus then coulombic force of attraction is called centripetal force.
c) When a satellite revolves around earth then the gravitational force of attraction is called centripetal force.




d) If a block is placed on disc and the block is rotating with the disc then the frictional force is called
centripetal force.



, When a car moves at a steady speed around an unbanked curve, the centripetal force keeping the
car on the curve comes from the static friction between the road and the ties.
It is static, rather than kinetic friction, because the tires are not slipping with respect to the radial
direction.
If the static frictional force is insufficient, given the speed and the radius of the turn, the car will
skid off the road.




static frictional
force



Ex. 1 A small body of mass tied with a string and kept on a smooth horizontal table is rotated over the table
in a circle of radius R with a constant speed of vo . Find the tension in the string.
Ex. 2 A conical pendulum of mass m and length L is suspended from the ceiling and the bob is rotated in a
horizontal circle such that it makes angle  with the vertical. Find the tension in the string and the
tangential speed of the bob.

Vertical Circular Motion of a Pendulum
A bob of mass m tied with a string of length R is suspended vertically. At the lowest position it is given a horizontal
velocity v0 so that it is whirled in a vertical circle.
From the FBD of the bob at any time t = t (see figure) C
Along radial direction :
T - mg cos  = m a aC
C
2 B aT
mv
 T - mg cos  = ......(i) 
R
Along tangential direction : mgsin
mg sin  = ma A mgcos


T
 aT V0 mg
= g sin 

Kinematically :
aT = - g sin 
dv
 gsin 
dt
dv d dv
 gsin  

  gsin 
d dt d
dv v
 gsin 
d R
v 
  v
0
vdv  Rg  sin d
0


 v= .............(ii)
v20 + 2Rg(cosθ - 1)

ANALYSIS OF VERTICAL CIRCULAR MOTION

Case 1 : 0   
 cos  ve
2
 To find whether T=0 is possible or not.

, Putting T = 0 in eq. (i) we get
 -mg cos   mv2/R  invalid equation
Therefore T = 0 is not possible.