Examples, exercises and solutions - Classical Mechanics & Special Relativity for Starters

Exercise 2 (Units based on physical constants ^[1] 🌶 🌶)

In physics, we assume that quantities like the speed of light ( $c$ ) and Newton’s gravitational constant ( $G$ ) have the same value throughout the universe, and are therefore known as physical constants. A third such constant from quantum mechanics is Planck’s constant ( $\hbar$ , $h$ an with a bar). In high-energy physics, people deal with processes that occur at very small length scales, so our regular SI-units like meters and seconds are not very useful. Instead, we can combine the fundamental physical constants into different basis values.

Combine $c$ , $G$ and $\hbar$ into a quantity that has the dimensions of length.
Calculate the numerical value of this length in SI units (this is known as the Planck length). You can find the numerical values of the physical constants in Section 16.
Similarly, combine $c$ , $G$ and $\hbar$ into a quantity that has the dimensions of energy (indeed, known as the Planck energy) and calculate its numerical value.

Exercise 3 (Reynolds numbers ^[1] 🌶 🌶)

Physicists often use dimensionless quantities to compare the magnitude of two physical quantities. Such numbers have two major advantages over quantities with numbers. First, as dimensionless quantities carry no units, it does not matter which unit system you use, you’ll always get the same value. Second, by comparing quantities, the concepts ‘big’ and ‘small’ are well-defined, unlike for quantities with a dimension (for example, a distance may be small on human scales, but very big for a bacterium). Perhaps the best known example of a dimensionless quantity is the Reynolds number in fluid mechanics, which compares the relative magnitude of inertial and drag forces acting on a moving object:

\mathrm{Re} = \frac{\text{inertial forces}}{\text{drag forces}} = \frac{\rho v L}{\eta}

(1)

where $\rho$ is the density of the fluid (either a liquid or a gas), $v$ the speed of the object, $L$ its size, and $\eta$ the viscosity of the fluid. Typical values of the viscosity are $1.0\;\mathrm{mPa}\cdot\mathrm{s}$ for water, $50\;\mathrm{mPa}\cdot\mathrm{s}$ for ketchup, and $1.0\; \mu\mathrm{Pa}\cdot\mathrm{s}$ for air.

Estimate the typical Reynolds number for a duck when flying and when swimming (you may assume that the swimming happens entirely submerged). NB: This will require you looking up or making educated guesses about some properties of these birds in motion. In either case, is the inertial or the drag force dominant?
Estimate the typical Reynolds number for a swimming bacterium. Again indicate which force is dominant.
Oil tankers that want to make port in Rotterdam already put their engines in reverse halfway across the North sea. Explain why they have to do so.
Express the Reynolds number for the flow of water through a (circular) pipe as a function of the radius $R$ of the pipe, the volumetric flow rate (i.e., volume per second that flows through the pipe) $Q$ , and the kinematic viscosity $\nu \equiv \eta / \rho$ .
For low Reynolds number, fluids will typically exhibit so-called laminar flow, in which the fluid particles all follow paths that nicely align (this is the transparent flow of water from a tap at low flux). For higher Reynolds number, the flow becomes turbulent, with many eddies and vortices (the white-looking flow of water from the tap you observe when increasing the flow rate). The maximum Reynolds number for which the flow is typically laminar is experimentally measured to be about 2300. Estimate the flow velocity and volumetric flow rate of water from a tap with a 1.0 cm diameter in the case that the flow is just laminar.

Exercise 5 (Moving a box 🌶)

A box is on a friction incline of 10°. It is pushed upward with a force $F_i$ for $\Delta t = 0.5$ s. It is then moving upward (inertia) but slows down due to gravity.

Below is a part of the python code. However, some essential elements of the code are missing indicated by (..).

Include the correct code and run it.
Explain the two graphs, highlighting all essential features of the graph by relating these to the given problem.
At what time is the acceleration 0? At what time is the box back at its origin?

The above context is not very realistic as friction is neglected. We, however, can include friction easily as it is given by $\vec{F_w} = \mu \vec{F_N}$ , with $\mu=0.05$ . Note that the direction of friction changes when the direction of the velocity changes!

Extend the code so that friction is included.

# Moving a box

## Importing libraries
import numpy as np
import matplotlib.pyplot as plt

## Constants
m = 2 #kg
F = 30 #N
g = 9.81 #m/s^2
theta = np.deg2rad(10) #degrees

## Time step
dt = 0.01 #s
t = np.arange(0, 10, dt) #s
t_F_stop = ..

## Initial conditions
x = np.zeros(len(t)) #m
v = np.zeros(len(t)) #m/s


## Loop to calculate position and velocity
for i in range(1, len(t)):
    if t[i] < ..:
        a = F/m - g*np.sin(theta)
    else:
        a = -g*np.sin(theta)
    v[i] = v[i-1] + ..
    x[i] = x[i-1] + v[i]*dt

## Plotting results
figs, axs = plt.subplots(1, 2, figsize=(10, 5)) 

axs[0].set_xlabel('Time (s)')
axs[0].set_ylabel('Velocity (m/s)')
axs[0].plot(t, v, 'k.', markersize=1)


axs[1].set_xlabel('Time (s)')
axs[1].set_ylabel('Position (m)')
axs[1].plot(t, x, 'k.', markersize=1)



plt.show()

1.6.2Solutions¶

Solution to Exercise 1

\begin{aligned} F &= k v^2\\ [\frac{\text{kgm}}{\text{s}^2}] &= [.][\frac{\text{m}^2}{\text{s}^2}] \Rightarrow [.]=[\frac{\text{kg}}{\text{m}}] \end{aligned}

(2)

Solution to Exercise 2

To be processed

Solution to Exercise 3

To be processed

Solution to Exercise 4

$=10^{-12}$
$=10^{21}$
$=10^{-3}$

Solution to Exercise 5

# Moving a box

## Importing libraries
import numpy as np
import matplotlib.pyplot as plt


part_4 = 1 # Turn to 0 for first part

## Constants
m = 2 #kg
F = 30 #N
g = 9.81 #m/s^2
theta = np.deg2rad(10) #degrees

mu = 0.02
F_N = m*g*np.cos(theta) #N

## Time step
dt = 0.01 #s
t = np.arange(0, 10, dt) #s
t_F_stop = 0.5

## Initial conditions
x = np.zeros(len(t)) #m
v = np.zeros(len(t)) #m/s


## Loop to calculate position and velocity
for i in range(0, len(t)-1):
    if t[i] < t_F_stop:
        a = F/m - g*np.sin(theta) - F_N*mu*np.where(v[i] != 0, np.sign(v[i]), 0)*part_4
    else:
        a = -g*np.sin(theta) - F_N*mu*np.where(v[i] != 0, np.sign(v[i]), 0)*part_4
    v[i+1] = v[i] + a*dt
    x[i+1] = x[i] + v[i]*dt

## Plotting results
figs, axs = plt.subplots(1, 2, figsize=(10, 5)) 

axs[0].set_xlabel('Time (s)')
axs[0].set_ylabel('Velocity (m/s)')
axs[0].plot(t, v, 'k.', markersize=1)


axs[1].set_xlabel('Time (s)')
axs[1].set_ylabel('Position (m)')
axs[1].plot(t, x, 'k.', markersize=1)



plt.show()

Solution to Exercise 6

# Simulation of a base jumper 

## Importing libraries
import numpy as np
import matplotlib.pyplot as plt

## Constants
A = 0.7 #m^2
m = 75 #kg
k = 0.37 #kg/m
g = 9.81 #m/s^2

## Time step
dt = 0.01 #s
t = np.arange(0, 12, dt) #s

## Initial conditions
z = np.zeros(len(t)) #m
v = np.zeros(len(t)) #m/s
z[0] = 300 #m

## Deploy parachute
A_max = 42.6 #m^2
t_deploy_start = 2 #s
dt_deploy = 3.8 #s

## Loop to calculate position and velocity
for i in range(0, len(t)-1):
    F = - m*g - k*A*abs(v[i])*v[i]  #N
    v[i+1] = v[i] + F/m*dt #m/s
    z[i+1] = z[i] + v[i]*dt #m
    # Check if the jumper is on the ground
    if z[i+1] < 0:
        break
    # Deploy parachute
    if t[i] > t_deploy_start and t[i] < t_deploy_start + dt_deploy:
        A += (A_max - A)/dt_deploy*dt 

## Plotting results
figs, axs = plt.subplots(1, 2, figsize=(10, 5)) 

axs[0].set_xlabel('Time (s)')
axs[0].set_ylabel('Velocity (m/s)')

axs[0].plot(t, v, 'k.', markersize=1, label='numerical solution')
axs[0].vlines(t_deploy_start, v[t==t_deploy_start],0, color='gray', linestyle='--', label='parachute deploy')

axs[0].legend()

axs[1].set_xlabel('Time (s)')
axs[1].set_ylabel('Position (m)')

axs[1].plot(t, z, 'k.', markersize=1)
axs[1].vlines(t_deploy_start, 150,300, color='gray', linestyle='--', label='parachute deploy')


plt.show()

Solution to Exercise 7

import numpy as np
import matplotlib.pyplot as plt

F = 49/3
m1 = 1
dt = 0.001
t = np.arange(0, 100, dt) # s

x1 = np.zeros(len(t)) # m
x1[0] = 3
y1 = np.zeros(len(t)) # m
vx = 0
vy = 7


for i in range(0, len(t)-1):
    ax = -F*(x1[i]-0)/np.sqrt(x1[i]**2 + y1[i]**2)
    ay = -F*(y1[i]-0)/np.sqrt(x1[i]**2 + y1[i]**2)
    vx = vx + ax*dt
    vy = vy + ay*dt
    x1[i+1] = x1[i] + vx*dt
    y1[i+1] = y1[i] + vy*dt


plt.figure(figsize=(4,4))
plt.plot(x1, y1, 'k.', markersize=1)
plt.xlabel('x (m)')
plt.ylabel('y (m)')
plt.show()

1.6Examples, exercises and solutions

1.6.1Exercises¶

1.6.2Solutions¶