# Chapter 13: Dynamics of a Particle¶

## Example 13.13-1, Page No 230¶

In [2]:
import math

#Initilization of variables
W=2 #lb
F=1.5 #lb
g=32.2 #ft/s**2

#Angles are with respect to the plane are,
# theta1=10 degrees & theta2=30 degrees
sintheta1=0.17
costheta1=0.99
sintheta2=0.5
costheta2=sqrt(3)*2**-1

#Calculations
#Now here the forces are considered as parallel and perpendicular to the plane
#Applying Newtond Principle
ax=(g/2)*(F*costheta1-(W*sintheta2)) #ft/s**2
N1=(2*costheta2-(F*sintheta1)) #lb

#result
print'The force on the particle is',round(N1,2),"lb"
print'The acceleration is',round(ax,2),"ft/s**2"

# The answer may wary due to decimal point descrepancy.

The force on the particle is 1.48 lb
The acceleration is 7.81 ft/s**2


## Example 13.13-2, Page No 231¶

In [9]:
import math

#Initilization of variables
m=5 #kg
s=12 #m
v=4 #m/s
vo=0 #m/s
g=9.8 #m/s**2
mu=0.25

#Calculations
#Using the kinematic equations of motion
a=(v**2-vo**2)*(2*s)**-1 #m/s**2
#Using Newtons Principle
N1=g*m #N
P=m*a+mu*N1 #N

#Result
print'The value of P is',round(P,1),"N"

The value of P is 15.6 N


## Example 13.13-3, Page No 232¶

In [11]:
import math

#Initilization of variables
m=2 #kg
vo=0 #m/s
v=3 #m/s
s=0.8 #m
# as theta=20 degrees,
sintheta=0.34
costheta=0.94
g=9.8 #m/s**2

#Calculations
N=m*g*costheta #N
a=(vo**2-v**2)*(2*s)**-1 #m/s**2
u=-((2*a)+(m*g*sintheta))/N
#Solving for return speed
#Symbol convention is different from textbook
a_ret=((m*g*sintheta)-(u*N))/2 #m/s**2
vf=sqrt((2*a_ret*s)) #m/s

#Result
print'The speed is',round(vf,1),"m/s"

The speed is 1.3 m/s


## Example 13.13-4, Page No 232¶

In [13]:
import math

#Initilization of variables
W=1800 #lb
r=2000 #ft
v=58.7 #ft/s
g=32.2 #ft/s**2

#Calculations
F=(W*v*v)/(g*r) #lb

#Result
print'The frictional force to be exerted is',round(F,1),"lb"

The frictional force to be exerted is 96.3 lb


## Example 13.13-7, Page No 234¶

In [15]:
import math
import numpy as np

# Initilization of variables
W=10 #lb
# as theta=30 degrees,
sintheta=2**-1
costheta=sqrt(3)*2**-1
l=2 #ft
w=10 #rev/min
g=32.2 # ft/s**2

# Calculations
r=l*costheta # ft
a_n=r*(((w*2*pi)/60)**2) #ft/s**2
#Applying Newtons Principle
#Solving by matrix method
A=np.array([[costheta,-sintheta],[sintheta,costheta]])
B=np.array([[(W*a_n)/g],[W]])
C=np.linalg.solve(A,B) #lb

#Result
print'The value of T is',round(C[0],2),"lb"

The value of T is 5.51 lb


## Example 13.13-8, Page No 235¶

In [16]:
import math

#Initilization of variables
m=4 #lb
v=6 #ft/s
r=2 #ft
# as theta1=40 degrees & theta2=20 degrees
sintheta1=0.64
costheta1=0.77
sintheta2=0.34
costheta2=0.94
g=32.2 #ft/s**2

#Calculations
a_n=v**2/r #ft/s**2
#Applying Newtons Principle
Fi=(m*a_n)/g #lb
#Solving by matrix method
A=np.array([[costheta1,costheta2],[sintheta1,-sintheta2]])
B=np.array([[m],[Fi]])
C=np.linalg.solve(A,B) #lb

#Result
print'The values are: T=',round(C[0],2),"lb",'and C=',round(C[1],2),"lb"

# The ans for C waries due to decimal point descrepancy.

The values are: T= 4.01 lb and C= 0.97 lb


## Example 13.13-10, Page No 237¶

In [18]:
import math

#Initilization of variables
m1=2 #kg
# as theta=20 degrees,
sintheta=0.34
m2=4 #kg
t=4 #s
g=9.8 #m/s**2
vo=0 #m/s

#Calculations
#Applying Newtons Principle
#Solving by matrix method
A=np.array([[1,-2],[1,4]])
B=np.array([[m1*g*sintheta],[m2*g]])
C=np.linalg.solve(A,B)
a=C[1] #m/s**2
v=vo+a*t #m/s

#Result
print'The velocity of 4kg mass is',round(v,1),"m/s"

The velocity of 4kg mass is 21.7 m/s


## Example 13.13-11, Page No 237¶

In [20]:
import math

#Initilization of variables
m_A=20 #lb
m_B=60 #lb
u=0.3 #coefficient of friction
t=4 #s
# as theta1=30 degrees & theta2=60 degrees,
sintheta1=2**-1
costheta1=sqrt(3)*2**-1
sintheta2=sqrt(3)*2**-1
costheta2=2**-1
g=32.2 #ft/s^2
vo=0 #ft/s

#Calculations
N1=m_A*costheta1 #lb
N2=m_B*costheta2 #lb
#Solving for T and a using matrix method
A=np.array([[1,-m_A/g],[-1,-m_B/g]])
B=np.array([[(m_A*sintheta1+u*N1)],[(-m_B*sintheta2+u*N2)]])
C=np.linalg.solve(A,B)
a=C[1] #ft/s**2
v=vo+a*t #ft/s

#Result
print'The velocity is',round(v,1),"ft/s"

# The ans in the textbook is incorrect.

The velocity is 44.7 ft/s


## Example 13.13-12, Page No 238¶

In [26]:
import math

#Initilization of variables
m_A=40 #kg
m_B=15 #kg
F=500 #N
g=9.8 #m/s**2
# as theta=30 degrees,
sintheta=2**-1
costheta=sqrt(3)*2**-1

#Calculations
m=m_A+m_B #kg
a=(F-m*g*sintheta)/(m) #m/s**2
#Summing forces parallel and perpendicular to the plane
#Simplfying equation (1) and (2)
Nb=m_B*g+(m_B*a*sintheta) #N
#Substituting this in eq(1)
u=-(m_B*g*costheta-(Nb*costheta))/(Nb*sintheta)

#Result
print'The value of u is',round(u,2)

The value of u is 0.31


## Example 13.13-13, Page No 239¶

In [28]:
import math

#Initilization of variables
P=70 #N
m_A=16 #kg
u_AH=0.25 #coefficient of friction between Block A and Horizontal Plane
m_B=4 #kg
u_BH=0.5 #coefficient of friction between Block B and Horizontal Plane
# as theta=10 degrees,
sintheta=0.17
costheta=0.98
g=9.8 #m/s**2

#Calculations
#Applying sum of forces to both the FBD's
#Solving by matrix method
A=np.array([[-costheta,-u_AH,-m_A,0],[-sintheta,1,0,0],[costheta,0,-m_B,-u_BH],[sintheta,0,0,1]])
B=np.array([[-P],[m_A*g],[0],[m_B*g]])
C=np.linalg.solve(A,B)

#Result
print'The Value of T is',round(C[0],1),"N"

# The ans waries due to decimal point descrepancy.

The Value of T is 20.7 N


## Example 13.13-14, Page No 239¶

In [35]:
import math

#Initilization of variables
# as theta=10 degrees
sintheta=0.1736
costheta=0.9848
v=10 #ft/s
v0=0 #ft/s
u=3**-1 #coefficient of friction
g=32.2 #ft/s**2

#Calculations
#Equations of motion for box are
#Simplfying the equations by sybstitution
a=((u*costheta)-(sintheta))*g #ft/s**2
#Time calculations
t=(v-v0)/a #s

#Result
print'The value of a is',round(a,2),"ft/s**2",'and the time required is',round(t),"seconds"

The value of a is 4.98 ft/s**2 and the time required is 2.0 seconds


## Example 13.13-15, Page No 240¶

In [39]:
import math

#Initilization of variables
g=9.8 #m/s**2

#Calculations
#Simplfying the equations we can solve for T2 and aA first to obtain the solution
#Solving by matrix method
A=np.array([[-1.5,-4],[-3.5,24]])
B=np.array([[-4*g],[-24*g]])
C=np.linalg.solve(A,B)
T2=C[0] #N
T1=T2/2 #N
T3=T2/2 #N
#Acceleration calculations
a1=1*g-T1 #m/s**2
a2=(2*g-T1)/2 #m/s**2
a3=(3*g-T3)/3 #m/s**2
a4=(4*g-T3)/4 #m/s**2
#Tension in fixed cord
T_f=2*T2 #N

#Result
print'The acceleration values are: a1=',round(a1),"m/s**2 (up)",',',round(a2,1),"m/s**2 (down)",',',round(a3,2),"m/s**2 (down)",',',round(a4,1),"m/s**2 (down) respectively."
print'The tension in the fixed cord is',round(T_f,1),"N"

The acceleration values are: a1= -9.0 m/s**2 (up) , 0.4 m/s**2 (down) , 3.53 m/s**2 (down) , 5.1 m/s**2 (down) respectively.
The tension in the fixed cord is 75.3 N


## Example 13.13-16, Page No 241¶

In [40]:
import math

#Initilization of variables
m1=14 #kg
m2=7 #kg
# as theta=45 degrees,
sintheta=sqrt(2)**-1
costheta=sqrt(2)**-1
u_1=4**-1 #coefficient of friction between mass 1 and plane
u_2=3*8**-1 #coefficient of friction between mass 2 and plane
g=9.8 #m/s**2

#Calculations
#The equations of motion for m1 are
N1=m1*g*costheta #N
F1=u_1*N1 #N
#The equations of motion for m2 are
N2=m2*g*costheta #N
F2=u_2*N2 #N
#Now to get T and a we solve using matrix method
A=np.array([[-1,-m1],[1,-m2]])
B=np.array([[-(m1*g*sintheta-F1)],[-(m2*g*sintheta-F2)]])
C=np.linalg.solve(A,B)

#Result
print'The Value of T is',round(C[0]),"N"

The Value of T is 4.0 N


## Example 13.13-19, Page No 244¶

In [41]:
import math

#Initilization of variables
W=12 #oz
k=2 #oz/in
M=0.34 #kg
K=22 #N/m
g=32.2 #ft/s**2

#Calculations
#Part(a)
a=(k*W*g)/16
b=W*16**-1
f=((2*pi)**-1)*((a/b)**0.5) #Hz for simplicity the numerator and denominator have been computed seperately as a and b
#Part(b)
F=((2*pi)**-1)*((K/M)**0.5) #Hz

#Result
print'The frequency in part (a) is',round(f,2),"Hz",'and in part(b) is',round(F,2),"Hz"

The frequency in part (a) is 1.28 Hz and in part(b) is 1.28 Hz


## Example 13.13-20, Page No 244¶

In [43]:
import math

#As the entire question is theoritical
#theta is directly computed
theta=arccos(2*3**-1)*(180/pi) #degrees

#result
print'The value of theta is',round(theta,1),"degrees"

The value of theta is 48.2 degrees


## Example 13.13-28, Page No 254¶

In [51]:
import math

#Initilization of variables
G=6.658*(10**-8)**-1 #cm**3/g.s**2
#Calculations
G1=G*((3.281*10**2)/((2.205*32.2**-1)*10**4)) #ft**3/slug-s**2
G2=G1 #ft**4/lb-s**4

#Result
print'The ans is',round(G2,2),"ft**4/lb-s**4"

# The ans waries slightly due to decimal point descrepancy.

The ans is 319004859.68 ft**4/lb-s**4


## Example 13.13-29, Page No 254¶

In [53]:
import math

#Initilization of variables
#Modifying the value of C without vo**2 in it
C=5000*5280
G=3.43*10**-8 #Gravatational Constant
M=4.09*10**23 #Mass of the Earth
a=5.31*10**8
#When the orbit is circular e=0
vo1=(a)**0.5 #ft/s
#When the orbit is parabolic e=1
vo2=((C*a+G*M)/C)**0.5 #ft/s

#Result
print'The value of vo1=',round(vo1),"ft/s",'is smaller than vo2=',round(vo2),"ft/s, hence the Satellite will enter a hyperbolic path and never return to earth."
#Decimal accuracy causes discrepancy in answers

The value of vo1= 23043.0 ft/s is smaller than vo2= 32594.0 ft/s, hence the Satellite will enter a hyperbolic path and never return to earth.


## Example 13.13-30, Page No 255¶

In [54]:
import math

#Initilization of variables
r=3940+500 #mi
phi=0 #degrees
vo=36000 #ft/s
C=4440*5280*vo
G=3.43*10**-8
M=4.09*10**23 #kg

#Calculations
e=((C*vo)/(G*M))-1

#Result
print'The value of e=',round(e,2),",hence the path is Hyperbolic"

The value of e= 1.17 ,hence the path is Hyperbolic


## Example 13.13-31, Page No 255¶

In [56]:
import math

#Initilization of variables
a=92.9*10**6 #mi
G=3.43*10**-8
T=365*24*3600 #s
c=5280

#Calculations
M=(4*pi**2*a**3*c**3)/(G*T**2) #slugs

#Result
print'The mass of the sun is',round(M,1),"slugs"

The mass of the sun is 1.36584467048e+29 slugs