# Chapter 4 : Motion in a Plane¶

## Example 4.1 , page : 69¶

In [1]:
# Importing module

import math

# Variable delcaration

v_r=35                   # Velocity vector  of the rain which falls vertically in m/s
v_w=12                   # Velocity vector of the wind blowing in east to west direction in m/s

# Calculation

R=math.sqrt(v_r**2+v_w**2)        # The magnitude of the resultant vector
tanθ=v_w/v_r
θ=math.degrees(math.atan(tanθ))   # The direction that R makes with the vertical

# Result

print("Therefore,the boy should hold his umbrella in the vertical plane at an angle of about",round(θ,0),"° with the vertical towards the east. ")

Therefore,the boy should hold his umbrella in the vertical plane at an angle of about 19.0 ° with the vertical towards the east.


## Example 4.2 , page : 71¶

In [2]:
# Calculation

# Let OP and OQ represent the two vectors A and B making an angle θ.
# Then, using the parallelogram method of vector addition, OS represents the resultant vector R such that R = A + B
# SN is normal to OP and PM is normal to OS.
# From the geometry of the figure, OS2 = ON2 + SN2 but ON = OP + PN = A + Bcosθ
# SN = Bsin θ and OS² = (A + Bcosθ )² + (Bsinθ )² or, R² = A² + B² + 2ABcosθ
# .i.e R = square root(A²+ B²+ 2ABcosθ)
# In ΔOSN, SN = OSsin α = Rsin α ,  and in ΔPSN,   SN = PSsinθ = Bsinθ
# Therefore, Rsinα  = Bsinθ  ..............eqn 1
# Similarly, PM = Asinα  = Bsinβ  .........eqn 2
# From eqns 1 and 2 we get,
# sin α = (B/R) sin θ or tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)

# Result

print("The magnitude of thr resultant vector is given by the equation, R = square root(A²+ B²+ 2ABcosθ)")
print("The direction of the resultant vector is given by the equation, tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)")

The magnitude of thr resultant vector is given by the equation, R = square root(A²+ B²+ 2ABcosθ)
The direction of the resultant vector is given by the equation, tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)


## Example 4.3 , page : 72¶

In [3]:
# Importing module

import math

# Variable declaration

v_b=25                   # Velocity of the motor boat racing towards north in km/h
v_c=10                   # Velocity of the water current in the direction of 60° east of south in km/h
θ=60                     # The angle in degree

# Calculation

# Using the parallelogram method of vector addition we can obtain the resultant vector
# We can obtain the magnitude of resultant vector using the Law of cosine
# We can obtain the direction of resultant vector using the Law of sines
φ=math.degrees(math.asin(sinφ))

# Result

print("The magnitude of the resultant vector =",round(R,0),"km/h")
print("The direction of the resultant vector =",round(φ,1),"°")

The magnitude of the resultant vector = 22.0 km/h
The direction of the resultant vector = 23.4 °


## Example 4.4 , page : 75¶

In [4]:
# Importing module

import math

# Variable declaration

t=1                     # Time in s

# Calculation

# The position vector of the particle is  r(t)=3.0ti+2.0t²j+5.0k
# On differentiating r(t) with respect to t we get the velocity vector, v(t)=3.0i+4.0tj
# On differentiating v(t) with respect to t we get the acceleration vector, a(t)=4.0j
V_x=3 # X component of v(t)
V_y=4 # Y component of v(t)
V=math.sqrt(V_x**2 + V_y**2)
tanθ=V_y/V_x
θ=math.degrees(math.atan(tanθ))

# Result

print("(a) The velocity vector, v(t) = 3.0i+4.0tj m/s")
print("    The acceleration vector, a(t) = 4.0j m/s²")
print("(b) Magnitude of v(t) at t=1 s =",V,"m/s")
print("    Direction of v(t) at t=1 s =",round(θ,0),"° with x-axis")

(a) The velocity vector, v(t) = 3.0i+4.0tj m/s
The acceleration vector, a(t) = 4.0j m/s²
(b) Magnitude of v(t) at t=1 s = 5.0 m/s
Direction of v(t) at t=1 s = 53.0 ° with x-axis


## Example 4.5 , page : 76¶

In [5]:
# Importing module

import sympy

# Variable declaration

x=84                     # X component of the position vector in m
t= sympy.symbols('t')

# Calculation

# Velocity vector is given as, v(t)= 5.0i  m/s
# Acceleration vector is given as, a(t)= 3.0i+2.0j  m/s²
# By the equation r(t)=v(t)+(a(t)t²)/2 we get the positon vector of the particle as follows
# r(t) = 5.0ti + 1.5t²i + 1.0t²j
t=round((max(sympy.solve(1.5*t**2 + 5*t -x,t))),0)
y=1.0*t**2
# Now the velocity vector can be obtained by differentiating r(t) with respect to t
# Then we get, v(t) = 5.0i +3ti+2tj m/s
v_x=5.0+3*t              # X component of v(t) at time = t
v_y=2*t                  # Y component of v(t) at time = t
V=math.sqrt(v_x**2 + v_y**2)

# Result

print("(a) The y-coordinate of the particle at the instant its x-coordinate is 84 m =",y,"m")
print("(b) Speed of the particle at the instant its x-coordinate is 84 m =",round(V,0),"m/s")

(a) The y-coordinate of the particle at the instant its x-coordinate is 84 m = 36.0 m
(b) Speed of the particle at the instant its x-coordinate is 84 m = 26.0 m/s


## Example 4.6 , page : 76¶

In [6]:
# Importing module

import math

# Variable declaration

v_r=35                   # Velocity vector of the rain which falls vertically in m/s
v_b=12                   # Velocity vector of the bicycle riding in east to west direction in m/s

# Calculation

v_rb=v_r - v_b
tanθ=v_b/v_r
θ=math.degrees(math.atan(tanθ))

# Result

print("Therefore, the woman should hold her umbrella at an angle of about",round(θ,0),"° with the vertical towards the west.")

Therefore, the woman should hold her umbrella at an angle of about 19.0 ° with the vertical towards the west.


## Example 4.7 , page : 78¶

In [7]:
# Variable declaration

v_0=1                    # For convenience, velocity at whch the projectile launched is assumed to be unity
θ_0=1                    # For convenience, angle at whch the projectile launched in degree is assumed to be unity

# Result

print("For a projectile launched with velocity v_0 at an angle θ_0 , the range is given by, R = (v_0²sin2θ_0)/g")
print("Now,for angles,(45°+α ) and (45°- α ), 2θ_0 is (90° + 2α) and (90° - 2α), respectively.")
print("The values of  sin (90° + 2 α ) and  sin (90° - 2α) are the same, equal to that of cos 2α.")
print("Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.")

For a projectile launched with velocity v_0 at an angle θ_0 , the range is given by, R = (v_0²sin2θ_0)/g
Now,for angles,(45°+α ) and (45°- α ), 2θ_0 is (90° + 2α) and (90° - 2α), respectively.
The values of  sin (90° + 2 α ) and  sin (90° - 2α) are the same, equal to that of cos 2α.
Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.


## Example 4.8 , page : 78¶

In [8]:
# Importing module

import math

# Variable declaration

# We choose the origin of the x-,and y- axis at the edge of the cliff and t = 0 s at the instant the stone is thrown
# Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the vertically upward direction
# The equations of motion are : x(t)=x_0 = v_0xt and y(t) = y_0+v_0yt+(1/2)a_y(t^2)

g=9.8                    # Acceleration due to gravity
x_0=0
y_0=0
v_oy=0
a_y=g
v_ox=15
y_t=490

# Calculation

# The stone hits the  ground when y(t) =  490 m ,i.e. 490  = (1/2)(9.8)t
t=math.sqrt((-y_t*-2)/a_y)
# The velocity components are v_x = v_ox  and v_y = v_oy - g t
v_x=v_ox
v_y=v_oy-(g*t)
V=math.sqrt(v_x**2+v_y**2)

# Result

print("The time taken by the stone to reach the ground =",t,"s")
print("The speed with which the stone hits the ground =",round(V,0),"m/s")

The time taken by the stone to reach the ground = 10.0 s
The speed with which the stone hits the ground = 99.0 m/s


## Example 4.9 , page : 79¶

In [9]:
# Importing module

import math

# Variable declaration

v_0=28                   # The initial velocity of the ball in m/s
θ=30                     # The angle of inclination of the ball above the horizontal in degree
g=9.8                    # Acceleration due to gravity in m/s²

# Calculation

# Result

print("The maximum height =",round(h_m,0),"m")
print("The time taken to return to the same level =",round(T_f,1),"s")
print("The distance from the thrower to the point where the ball returns to the same level =",round(R,0),"m")

The maximum height = 10.0 m
The time taken to return to the same level = 2.9 s
The distance from the thrower to the point where the ball returns to the same level = 69.0 m


## Example 4.10 , page : 81¶

In [10]:
# Importing module

import math

# Variable declaration

R=12                     # Radius of the circular groove in cm
n=7                      # Total number of revolutions
t=100                    # Time taken for all 7 revolutions

# Calculation

T=t/n                    # Time taken for one revolution
#(a)
w=2*math.pi/T            # Angular speed in rad/s
v=w*R                    # Linear speed in cm/s
a=pow(w,2)*R             # Magnitude of acceleration in cm/s²

# Result

print("(a) Angular speed =",round(w,2),"rad/s")
print("    Linear speed =",round(v,1),"cm/s")
print("(b) Since the direction changes continuously, acceleration here is not a constant vector")
print("    Magnitude of acceleration =",round(a,1),"cm/s²")

(a) Angular speed = 0.44 rad/s
Linear speed = 5.3 cm/s
(b) Since the direction changes continuously, acceleration here is not a constant vector
Magnitude of acceleration = 2.3 cm/s²