# Example 15.2 Components of Acceleration¶

In [1]:
# Initialization of variables
v_1=72*(1000/3600) # m/s # initial speed of the car
v_2=36*(1000/3600) # m/s # speed of the car after 10 seconds
t=10 # seconds
# Calculations
A_n=v_1**2/r # m/s**2 # normal component of acceleration
A_t=0 # since dv/dt=0 # tangential component of acceeration
delv=v_1-v_2
delt=t-0
a_t=delv/delt # m/s**2 # tangential component of deceleration after the brakes are applied
a_n=v_1**2/r # m/s**2 # normal component of deceleration  after the brakes are applied
# Results
print('The normal component of acceleration is %f m/s**2'%A_n)
print('The tangential component of acceleration is %f m/s**2'%A_t)
print('The normal component of deceleration is %f m/s**2'%a_n)
print('The tangential component of deceleration is %f m/s**2'%a_t)

The normal component of acceleration is 2.000000 m/s**2
The tangential component of acceleration is 0.000000 m/s**2
The normal component of deceleration is 2.000000 m/s**2
The tangential component of deceleration is 1.000000 m/s**2


# Example 15.3 Components of Acceleration¶

In [3]:
import math
#Initialization of variables
a_t=0.6 # m/s**2 # tangential acceleration
a=0.75 # m/s**2 # total acceleration attained by the car
# Calculations
a_n=math.sqrt(a**2-a_t**2) # m/s**2
v=math.sqrt(a_n*r) # m/s
# Using v=u+a*t
u=0
t=v/a_t # seconds
# Now using v**2-u**2=2*a*s
s=v**2/(2*a_t) # m
# Results
print('The distance  traveled by the car is %f m'%s)
print('The time for which the car travels is %f seconds'%t)

The distance  traveled by the car is 93.750000 m
The time for which the car travels is 17.677670 seconds


# Example 15.5 Motion of a particle on a curved frictionless path¶

In [2]:
from __future__ import division
import math
# Initialization of variables
v=10 # m/s # speed of the car
t=15 # seconds
# Calculations
omega=(v/r) # radian/seconds # angular velocity of the car
# Velocity in x & y direction is given by eq'n
v_x=omega*r*(math.sin((omega*(180/math.pi)*t)*math.pi/180)) # m/s # value of v_x is -ve but we consider it to be +ve for calculations
v_y=omega*r*(math.cos((omega*(180/math.pi)*t)*math.pi/180)) # m/s
# Acceleration in x & y direction is given by
a_x=omega**2*r*(math.cos((omega*(180/math.pi)*t)*math.pi/180)) # m/s**2 # value of a_x is -ve but we consider it to be +ve for calculations
a_y=omega**2*r*(math.sin((omega*(180/3.14)*t)*math.pi/180)) # m/s**2 # value of a_y is -ve but we consider it to be +ve for calculations
a=math.sqrt(a_x**2+a_y**2) # m/s**2 # total acc
phi=math.degrees(math.atan(a_y/a_x)) # degrees # direction of acceleration
# Components in tangential and normal directions
# Velocity
v_n=0 # m/s
v_t=v # m/s
# Acceleration
a_n=v**2/r # m/s**2 # normal acc
a_t=0 # tangential acc
# angular position of the car after 15 sec
theta=omega*(180/math.pi)*t # degrees
# Results
print('The component of velocity in X direction (v_x) is %f m/s'%v_x)
print('The component of velocity in Y direction (v_y) is %f m/s'%v_y)
print('The component of acceleration in X direction (a_x) is %f m/s**2'%a_x)
print('The component of acceleration in Y direction (a_y) is %f m/s**2'%a_y)
print('The total acceleration is %f m/s**2 and its direction is %f degrees'%(a,phi))
print('The normal acceleration is %f m/s**2 and tangential acceleration is %f m/s**2'%(a_n,a_t))

The component of velocity in X direction (v_x) is 6.816388 m/s
The component of velocity in Y direction (v_y) is 7.316889 m/s
The component of acceleration in X direction (a_x) is 0.365844 m/s**2
The component of acceleration in Y direction (a_y) is 0.340959 m/s**2
The total acceleration is 0.500095 m/s**2 and its direction is 42.983499 degrees
The normal acceleration is 0.500000 m/s**2 and tangential acceleration is 0.000000 m/s**2


# Example 15.6 Motion of a particle on a curved frictionless path¶

In [11]:
from __future__ import division
import math
# Initialization of variables
t=1 # seconds
# Calculations
# From the equations of r and theta given we find 1st & 2nd derative and substitute t=1sec Here we consider the 1st derative as r_1 & theta_1 and so on...
r=(1.25*t**2)-(0.9*t**3) # m
r_1=(1.25*(2*t))-(0.9*(3*t**2)) # m/s
r_2=2.5-(0.9*3*(2*t)) # m/s**2
# Velocity of collar P
v_r=r_1 # m/s
v_theta=r*theta_1 # m/s
v=math.sqrt(v_r**2+v_theta**2) # m/s
alpha=math.degrees(math.atan(v_theta/v_r)) # degree
# Acceleration of the collar P
a_r=r_2-(r*theta_1**2) # m/s**2
a_theta=(r*theta_2)+(2*r_1*theta_1) # m/s**2
a=math.sqrt(a_r**2+a_theta**2) # m/s**2
beta=math.degrees(math.atan(a_theta/a_r)) # degree
# Acceleration of collar P relative to the rod. Let it be a_relative
a_relative=r_2 # m/s**2 # towards O
# Calculations
print('The velocity of the collar is %f m/s'%v)
print('The accelaration of the collar is %f m/s**2'%a)
print('The acceleration of the collar relative to the rod is %f m/s**2'%a_relative)

The velocity of the collar is 1.117599 m/s
The accelaration of the collar is 6.674415 m/s**2
The acceleration of the collar relative to the rod is -2.900000 m/s**2


# Example 15.7 Components of motion¶

In [12]:
from __future__ import division
import math
#Initialization of variables
# Consider the eq'ns of motion from the book
# The notations have been changed for the derivatives of r & theta
# (1) At t=0 s
theta_0=0
theta_2=0
r_0=0
r_1=10 # cm/s
r_2=0
# At t=0.3 s
t=0.3 # sec
theta2=0
r=10*t # cm
r1=10 # cm/s
r2=0
# (i)
#Velocity
v_r=r_1 # cm/s
v_theta=r_0*theta_1
v=math.sqrt(v_r**2+v_theta**2) # cm/s
# Acceleration
a_r=r_2-(r_0*theta_1**2) # cm/s**2
a_theta=(r_0*theta_2)+(2*r_1*theta_1) # cm/s**2
a=math.sqrt(a_r**2+a_theta**2) # cm/s**2
# (ii)
# Velocity
V_R=r1 # cm/s
V_theta=r*theta1 # cm/s
V=math.sqrt(V_R**2+V_theta**2) # cm/s
# Acceleration
A_r=r2-(r*theta1**2) # cm/s**2
A_theta=(r*theta2)+(2*r1*theta1) # cm/s**2
A=math.sqrt(A_r**2+A_theta**2) # cm/s**2
# Results
print('The velocity and the acceleration of the partice at t=0 s is %f cm/s & %f cm/s**2'%(v,a))
print('The velocity and the acceleration of the partice at t=0.3 s is %f cm/s & %f cm/s**2'%(V,A))

The velocity and the acceleration of the partice at t=0 s is 10.000000 cm/s & 125.663706 cm/s**2
The velocity and the acceleration of the partice at t=0.3 s is 21.337895 cm/s & 172.679692 cm/s**2


# Example 15.9 Equations of dynamic equilibrium¶

In [13]:
from __future__ import division
import math
# Calculations
# Tension in the wire before it is cut
T_ab=1/((2.747*0.643)+(0.766)) # From eqn's 1 & 2..Here T_ab is multiplied with W (i.e weight of small ball)
T_AB=math.cos(40*math.pi/180) # Tension in the wire after the wire is cut. Again T_AB is multiplied with W.
# Results
print('The tension in the wire before and after it is cut is respectively %f W & %f W'%(T_ab,T_AB))

The tension in the wire before and after it is cut is respectively 0.394895 W & 0.766044 W


# Example 15.10 Equations of dynamic equilibrium¶

In [14]:
from __future__ import division
import math
#Initialization of variables
mu_a=0.40 # coefficient of friction under block A
n=40 # r.p.m # speed of rotation of frame
W_A=120 # N # weight of block A
W_B=80 # N # weight of block B
r_1=1.2 # m # distance between W_A & axis of rotation
r_2=1.6 # m # distance between W_B & axis of rotation
g=9.81 # m/s**2
# Calculations
# Consider the F.B.D of block A
N=W_A # N # sum F_y=0
a_n=omega**2*r_1 # m/s**2
T=((W_A/g)*a_n)-(mu_a*W_A) # N
# Now consider the F.B.D of block B
A_n=omega**2*r_2 # m/s**2
N_1=(W_B/g)*A_n # N # sum F_x=0
mu=(T-W_B)/N_1 # sum F_x=0
# Results
print('The coefficient of friction of block B is %f'%mu)

The coefficient of friction of block B is 0.565897


# Example 15.12 Motion of particle on curved frictionless path¶

In [3]:
#Initialization of variables
W=10000 # N # Weight of the locomotive
# Calculations
# Consider the various derivations given in the textbook
R_max=W/20 # N # eq'n for  max reaction
# The position of occurence of maximum thrust cannot be defined here. Refer textbook for the answer
# Results
print('The maximum lateral thrust is %f N'%R_max)

The maximum lateral thrust is 500.000000 N


# Example 15.13 Motion of a particle on curved frictionless path¶

In [16]:
from __future__ import division
import math
# Initialization of variables
W=10 # N # Weight of the ball
# Calculations
# consider the eq'n derived to find the reaction, given as
R=W*(1+((2*math.pi**2)/9)) # N
# Results
print('The value of the reaction is %f N'%R)

The value of the reaction is 31.932454 N


# Example 15.15 Motion of a particle in a curved frictionless path¶

In [17]:
from __future__ import division
import math
#Initialization of variables
P=50 # N # Weight of ball P
Q=50 # N # Weight of ball Q
R=100 # N # Weight of the governing device
l=0.3 # m # length of each side
theta=30 # degree
g=9.81 # m/s**2 # acc due to gravity
# Calculations
# Consider the respective F.B.D
r=l*math.sin(theta*math.pi/180) # m # Radius of circe
# On solving eqn's 1,2 &3 we get the value of v as,
v=math.sqrt(((Q+R)*g*r)/((math.sqrt(3))*Q)) # m/s
# But the eq'n v=omega*r we get the value of N as,
N=(60*v)/(2*math.pi*r) # r.p.m
# Results
print('The speed of rotation is %f r.p.m'%N)

The speed of rotation is 101.634363 r.p.m


# Example 15.16 Motion of a particle in a curved frictionless path¶

In [18]:
from __future__ import division
import math
# Initilization of variables
Q=20 # N # Weight of the governor device
W=10 # N # Weight of the fly balls
theta=30 # degree # angle between the vertical shaft and the axis AB
l=0.2 # m # length of the shaft
g=9.81 # m/s**2 # acc due to gravity
# Calculations
# Consider the respective F.B.D
# Radius of the circle is given as,
r=Q*math.sin(theta*math.pi/180)*(10**-2) # m
# Solving eq'n 1 & 2 for v. The eq'n for v is given as,
v=math.sqrt(((W*l*0.5)+(0.05*Q))/((W*0.2*math.sqrt(3))/(2*g*r))) # m/s
# But, v=r*omega=2*pi*N*r/60. From this eq'n we get N as,
N=(v*60)/(2*math.pi*r) # r.p.m.
# Results
print('The speed of the fly-balls is %f r.p.m'%N)

The speed of the fly-balls is 101.634363 r.p.m


# Example 15.18 Motion of vehicles on leveled and banked roads¶

In [19]:
from __future__ import division
import math
#Initialization of variables
mu=0.15 # coefficient of friction between the wheels and the road
g=9.81 # m/s**2 # acc due to gravity
# Calculations
# The eq'n fo max speed of the vehicle without skidding is
v=math.sqrt(mu*g*r) # m/s
# The angle theta made with the vertical while negotiating the corner is
theta=math.degrees(math.atan(v**2/(g*r))) # degree
# Results
print('The maximum speed with which the vehicle can travel is %f m/s'%v)
print('The angle made with the vertical is %f degree'%theta)

The maximum speed with which the vehicle can travel is 8.577587 m/s
The angle made with the vertical is 8.530766 degree


# Example 15.19 Motion of vehicles on leveled and banked roads¶

In [20]:
from __future__ import division
import math
#Initialization of variables
v=100*(1000/3600) # m/s # or 100 km/hr
g=9.81 # m/s**2 # acc due to gravity
# Calculations
# The angle of banking is given by eq'n,
theta=math.degrees(math.atan((v**2)/(g*r))) # degree
# Results
print('The angle of banking of the track is %f degree'%theta)

The angle of banking of the track is 17.464605 degree


# Example 15.20 Motion of vehicles on leveled and banked roads¶

In [21]:
from __future__ import division
import math
#Initialization of variables
W=10000 # N # Weight of the car
v=10 # m/s # speed of the car
h=1 # m # height of the C.G of the car above the ground
b=1.5 # m # distance between the wheels
g=9.81 # m/s**2 # acc due to gravity
# Calculations
# The reactions at the wheels are given by te eq'ns:
R_A=(W/2)*(1-((v**2*h)/(g*r*b))) # N # Reaction at A
R_B=(W/2)*(1+((v**2*h)/(g*r*b))) # N # Reaction at B
# The eq'n for max speed to avoid overturning on level ground is,
v_max=math.sqrt((g*r*(b/2))/(h)) # m/s
# Results
print('The reaction at Wheel A (R_A) is %f N'%R_A)
print('The reaction at Wheel B (R_B) is %f N'%R_B)
print('The maximum speed at which the vehicle can travel without the fear of overturning is is %f m/s'%v_max)

The reaction at Wheel A (R_A) is 4660.210669 N
The reaction at Wheel B (R_B) is 5339.789331 N
The maximum speed at which the vehicle can travel without the fear of overturning is is 27.124712 m/s


# Example 15.21 Motion of vehicles on leveled and banked roads¶

In [22]:
from __future__ import division
import math
#Initialization of variables
W=1 # N # Weight of the bob
theta=8 # degree # angle made by the bob with the vertical
r=100 # m # radius of the curve
g=9.81 # m/s**2 # acc due to gravity
# Calculations
# from eq'n 1 & 2 we get v as,
v=(math.sqrt(g*r*math.tan(theta*math.pi/180)))*(3600/1000) # km/hr
T=W/math.cos(theta*math.pi/180) # N # from eq'n 2
# Results
print('The speed of the cariage is %f km/hr'%v)
print('The tension in the chord is %f N'%T)

The speed of the cariage is 42.270586 km/hr
The tension in the chord is 1.009828 N