# CHAPTER07 : CIRCULAR MOTION¶

## Example E1 : Pg 101¶

In [1]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.1
# calculation of the angular velocity

# given data
v=10.# linear speed(in m/s)
r=20.*10.**-2.# radius(in cm) of the circle

# calculation
w=v/r# formula of angular velocity


the angular velocity is 50.0 rad/s


## Example E2 : Pg 101¶

In [2]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.2
# calculation of the angular acceleration

# given data
v1=5.# initial speed(in m/s)
v2=6.# final speed(in m/s)
dt=2.# change in time(in s)
r=20.*10.**-2.# radius(in cm) of the circle

# calculation
at=(v2-v1)/dt# formula of tangential acceleration
alpha=at/r# formula of angular acceleration


the angular accleration is 2.5 rad/s**2


## Example E3 : Pg 102¶

In [3]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.3
# calculation of the magnitude of linear acceleration
# given data
import math
from math import pi
t=4.# time(in s) taken

# calculation
d=2*math.pi*r# distance covered
v=d/t# linear speed
a=(v*v)/r

print'the linear acceleration is',a,'m/s**2'

the linear acceleration is 0.246740110027 m/s**2


## Example E4 : Pg 103¶

In [4]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.4
# calculation of the value of radial and tangential acceleration

# given data
t=3.# time(in s)
r=20.*10.**-2.# radius(in cm) of the circle

#function v1=f(t1)
#  v1=2*t1
#endfunction

# calculation
v=6.;#f(t)
at=2.;#derivative(f,t)# tangential acceleration

print'the value of radial acceleration is',ar,'m/s**2'
print'the value of tangential acceleration is',at,'m/s**2'

the value of radial acceleration is 180.0 m/s**2
the value of tangential acceleration is 2.0 m/s**2


## Example E5 : Pg 103¶

In [5]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.5
# calculation of the normal contact force by the side wall of the groove
# given data
import math
r=25.*10.**-2.# radius(in m) of the cirlce
m=.1# mass(in kg) of the block
t=2.# time(in s) taken by the block

# calculation
v=2.*math.pi*r/t# speed of the block
a=(v*v)/r# acceleration of the block
N=m*a# newton's second law

print'the normal contact force by the side wall of the groove is',N,'N'

the normal contact force by the side wall of the groove is 0.246740110027 N


## Example E6 : Pg 103¶

In [6]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.6
# calculation of the speed of vehicle on the turn
# given data
r=10.# radius(in m) of the turn
theta=10.# angle(in degree) of the bank
g=9.8# gravitational acceleration(in m/s**2) of the earth
# calculation
v=4.16;#sqrt(r*g*tand(theta))# since tand(theta) = (v*v)/(r*g)
print'for normal contact force providing the necessary centripetal force,the speed of vehicle on the turn is',v,'m/s'

for normal contact force providing the necessary centripetal force,the speed of vehicle on the turn is 4.16 m/s


## Example E7 : Pg 107¶

In [7]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.7
# calculation of the weight of the body if spring balance is shifted to the equator
# given data
import math
W=98.# weight(in N) of the body at north pole
R=6400.*10.**3.# radius(in m) of the earth
g=9.8# gravitational acceleration(in m/s**2) of the earth

# calculation
m=W/g# formula of weight
w=(2.*math.pi)/(24.*60.*60.)# angular speed of the earth
We=W-(m*w*w*R)#   since We = W - (m*w*w*R)

print'the weight of the body if spring balance is shifted to the equator is',We,'N'

the weight of the body if spring balance is shifted to the equator is 97.6615362002 N


# WORKED EXAMPLES¶

## Example E1w : Pg 107¶

In [14]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.1w
# calculation of the maximum speed the car can take on the turn without skidding
# given data
import math
R=45.# radius(in m) of the turn
mus=2.0# coefficient of static friction between the tyre and the road
g=10.# gravitational acceleration(in m/s**2) of the earth

# calculation
# considering forces in vertical and horizontal directions an dpplying Newton's law we get
# fs = M*v*v...........................(1)
# by equation of limiting friction,we get
# fs = mus*N = mus*M*g...............(2)
# from above equations we get
v=math.sqrt(mus*g*R)

print'the maximum speed the car can take on the turn without skidding is',v,'m/s','or',(v*10**-3*60*60),'km/hr'

the maximum speed the car can take on the turn without skidding is 30.0 m/s or 108.0 km/hr


## Example E2w : Pg 107¶

In [9]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.2w
# calculation of the value of angle of banking

# given data
r=600.# radius(in m) of the track
v=180.*10.**3./(60.*60.)# speed(in m/s) of the car
g=10.# gravitational acceleration(in m/s**2) of the earth

# calculation
# for vertical direction
# N*cosd(theta) = M*g.........................(1)
# for horizontal direction
# N*sind(theta) = M*v*v/r.....................(2)
# from above equations,we get
theta=22.6;#atand((v*v)/(r*g))

print'the value of angle of banking is',theta,'degree'

the value of angle of banking is 22.6 degree


## Example E4w : Pg 108¶

In [10]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.4w
# calculation of the value of elongation of the spring

# given data
k=100.# spring constant(N/m) of the given spring
l0=.5# natural length(in m) of the string
m=.5# mass(in kg) of the particle
w=2.# angualr velocity(in rad/s) of the mass

# calculation
# from the equation of horizontal force
# k*l = m*v*v/r = m*w*w*r = =m*w*w*(l0+l)........................(1)
# from above equation we get
l=(m*w*w*l0)/(k-(m*w*w))

print'the value of elongation of the spring is',l,'m or',l*100,'cm'

the value of elongation of the spring is 0.0102040816327 m or 1.02040816327 cm


## Example E7w : Pg 109¶

In [11]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.7w
# calculation of the value of force exerted by the air on the plane

# given data
v=900.*10.**3./(60.*60.)# speed(in m/s) of the fighter plane
r=2000.# radius(in m)of the vertical circle
M=16000.# mass(in kg)
g=9.8# gravitational acceleration(in m/s**2) of the earth

# calculation
# from Newton's second law
# F-M*g = M*v*v/r
# from above equation,we get
F=M*(g+(v*v/r))

print' the force exerted by the air,on the plane in upward direction is',F,'N'

 the force exerted by the air,on the plane in upward direction is 656800.0 N


## Example E8w : Pg 109¶

In [12]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.8w
# calculation of the angular speed of rotation

# given data
L=20.*10.**-2.# length(in m) of the rod = length(in m)of the string
theta=30.# angle(in degree) made by the string with the vertical
g=10.# gravitational acceleration(in m/s**2) of the earth

# calculation
# applying Newton's second law
# T*sind(theta) = m*w*w*L*(1+sind(theta)).............(1)
# applying Newton's first law in vertical direction
# T*cosd(theta) = m*g.................................(2)
# from above equations,we get
# tand(theta)=((w*w*L*(1+sind(theta)))/g).............(3)
w=4.39;#sqrt((g*tand(theta))/(L*(1+sind(theta))))

print'the angular speed of rotation is',w,'rad/s'

the angular speed of rotation is 4.39 rad/s


## Example E10w : Pg 110¶

In [13]:
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 7.10w
# calculation of the minimum speed at which floor may be removed
# given data
import math
r=2.# radius(in m) of the rotor
mus=0.2# coefficient of static friction between the wall and the person
g=10.# gravitational acceleration(in m/s^2) of the earth

# calculation
# by applying Newton's second law for horizontal direction
# fs = m*g...................................(1)
# by limiting friction
# mus*N = m*g or mus*m*v*v/r = m*g...........(2)
# from above equations,we get
v=math.sqrt(r*g/mus)

print'the minimum speed at which floor may be removed is',v,'m/s'

the minimum speed at which floor may be removed is 10.0 m/s