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

import numpy as np
import matplotlib.pyplot as plt
import ipywidgets as widgets
from ipywidgets import interact

# --- Parameters ---
x0 = 0.0
t_stop = 2.0
dt = 0.02
t = np.arange(0, t_stop + dt, dt)

def run_animation(v0, F, m):
    # --- Derived motion ---
    x = x0 + v0 * t + 0.5 * F / m * t**2
    v = v0 + F / m * t
    E_kin = 0.5 * m * v**2
    W = F * (x - x0)

    # --- Plot setup ---
    plt.clf()
    
    fig, axs = plt.subplots(2, 1, figsize=(15, 10))
    
    ax_position = axs[0]
    ax_ekin = axs[1]

    ax_position.plot(t,x,'k-')
    ax_position.set_xlim(0, t_stop)
    ax_position.set_ylim(min(x)-1, max(x)+1)
    ax_position.set_title("Position vs Time")
    ax_position.set_ylabel('$x$(m)')

    ax_ekin.plot(t,E_kin,'k-',label='$E_{kin}$')
    ax_ekin.plot(t,W,'r-',label='$W$')
    ax_ekin.set_xlim(0, t_stop)
    ax_ekin.set_ylim(0, max(E_kin)*1.1)
    ax_ekin.set_title("Kinetic Energy vs Time")
    ax_ekin.set_ylabel('$E$(J)')
    

    
# --- Interact control panel ---
interact(run_animation,
         v0=widgets.FloatSlider(min=0, max=10, step=0.5, value=5.0, description="v₀ (m/s)"),
         F=widgets.FloatSlider(min=-5, max=5, step=0.5, value=6.5, description="F (N)"),
         m=widgets.FloatSlider(min=0.5, max=5.0, step=0.1, value=1.0, description="m (kg)"))

Loading...

3.8.2Answers set 1¶

Solution to Exercise 1

a. Show $\vec{\nabla} \times m\vec{g} = 0$
$\vec{\nabla} \times m\vec{g} = 0$ ? How to compute it? For Cartesian coordinates there is an easy to remember rule:

\vec{\nabla} \times \vec{F} = \begin{vmatrix}\hat{x}&\hat{y}&\hat{z}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\F_x&F_y&F_z\end{vmatrix}

(1)

If we chose our coordinates such that $\vec{g} = -g \hat{z}$ we get:

\vec{\nabla} \times \vec{F}_g = \begin{vmatrix}\hat{x}&\hat{y}&\hat{z}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\0&0& -mg\end{vmatrix} = 0

(2)

Thus $\vec{F}_g$ is conservative.

b. Find a $V$ that satisfies $-m\vec{g} = -\vec{\nabla} V$
Does $-m\vec{g} = -\vec{\nabla} V$ have a solution for V? Let’s try, using the same coordinates as above.

\begin{split} -\vec{\nabla}V &= - m\vec{g} \Rightarrow \\ \frac{\partial V}{\partial x} &= 0 \rightarrow V(x,y,z) = f(y,z) \\ \frac{\partial V}{\partial y} &= 0 \rightarrow V(x,y,z) = g(x,z) \\ \frac{\partial V}{\partial z} &= mg \rightarrow V(x,y,z) = mgz + h(x,y) \end{split}

(3)

f,g,h are unknown functions. But all we need to do, is find one $V$ that satisfies $-m\vec{g} = -\vec{\nabla} V$ .

So, if we take $V(x,y,z) = mgz$ we have shown, that gravity in this form is conservative and that we can take $V(x,y,z) = mgz$ for its corresponding potential energy.

By the way: from the first part (curl F = 0), we know that the force is conservative and we know that we could try to find V from

\begin{split} V(x,y,z) &= -\int_{ref} m\vec{g} \cdot d\vec{r} = \int_{ref} mg \hat{z} \cdot d\vec{r} \\ &= \int_{ref} mg dz = mgz + const \end{split}

(4)

Solution to Exercise 3

Click for the solution Friction Not Conservative.

Solution to Exercise 4

Click for the solution Conservative Force.

Solution to Exercise 5

Click for the solution Non-Conservative Force.

Solution to Exercise 6

Click for the solution Potential energy & Force.

Solution to Exercise 7

Click for the solution Force Field.

Solution to Exercise 8

Click for the solution Sinusoidal Force Field.

3.8.3Exercise set 2¶

Exercise 12 (Work done dragging a board ^[1] 🌶 🌶)

A uniform board of length $L$ and mass $M$ lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is $\mu_1$ , and in region 2, the coefficient is $\mu_2$ . Our objective is to find the net work $W$ done by friction in pulling the board directly from region 1 to region 2, under the assumption that the board moves at constant velocity.

Suppose that at some point during the process, the right edge of the board is a distance $x$ from the boundary, as shown. When the board is at this position, what is the magnitude of the force of friction acting on the board, assuming that it’s moving to the right? Express your answer in terms of all relevant variables ( $L$ , $M$ , $g$ , $x$ , $\mu_1$ , and $\mu_2$ ).
As we have seen, when the force is not constant, you can determine the work by integrating the force over the displacement, $W = \int F(x) \mathrm{d}x$ . Integrate your answer from (1) to get the net work you need to do to pull the board from region 1 to region 2.

Exercise 14 (🌶 🌶)

A hockey puck ( $m = 160$ gram) is hit and slides over the ice-floor. It starts at an initial velocity of $10 \mathrm{m/s}$ . The hockey puck experiences a frictional force from the ice that can be modeled as $-\mu F_g$ with $F_g$ the gravitational force on the puck. The friction force eventually stops the motion of the puck. Then the friction is zero (otherwise it would accelerate the puck from rest to some velocity :smiley: ). Constant $\mu = 0.01$ .

Set up the equation of motion (i.e. N2) for $m$ .
Solve the equation of motion and find the trajectory of the puck.
Calculate the amount of work done by gravity by solving the integral $W_{12} = \int_1^2 \vec{F} \cdot d\vec{r}$ .
Show that the amount of work calculated is indeed equal to the change in kinetic energy.
Solve this exercise using IDEA.

Exercise 17 (Work by a lineair force 🌶)

A point particle of mass $m=2\mathrm{kg}$ is at $t=0$ at position $x=0$ . It has initial velocity $v_0$ . From $t=0$ to $t_{stop}=4$ s it is under the influence of a force $F(x)$ that linearly increases with the position: $F(x) = k x$ with $k>0$ . This is a 1D problem.

The top graph show the position of the particle. The bottom graph shows the Work done on the particle by the force and the kinetic energy of the particle.

Analyse this situation and calculate the work done by the force at any time. Is the work done in this case always sufficient to account for the change in kinetic energy? What does it mean if the work is positive or negative?

Are the red ( $W$ ) line and the green ( $E_{kin}$ ) parallel? What does that mean?

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
from IPython.display import HTML

#  Inputs 
x0 = float(input('Initial position (m) from 0 to 10'))        # Initial position (m)
v0 = float(input('Initial velocity (m/s) from -10 to 10'))        # Initial velocity (m/s)
k = float(input('Spring constant (N/m) from 0.1 to 1'))       # Spring constant (N/m)
m = 2.0        # Mass (kg)

#  Derived 
omega = np.sqrt(k / m)
t_stop = 4.0
dt = 0.02
t_values = np.arange(0, t_stop, dt)

#  Analytical solution for x(t) 
C1 = 0.5 * x0 + 0.5 * v0 / omega
C2 = 0.5 * x0 - 0.5 * v0 / omega
x = C1 * np.exp(omega * t_values) + C2 * np.exp(-omega * t_values)

# Velocity is not needed directly as E_kin = total energy - work (from energy conservation)
W = 0.5 * k * (x**2 - x0**2)
v_squared = v0**2 + 2 * W / m
E_kin = 0.5 * m * v_squared

#  Plot setup 
fig, axs = plt.subplots(2, 2, figsize=(10, 10))
ax_block = axs[0, 0]
ax_pos = axs[1, 0]
ax_ekin = axs[0, 1]
ax_work = axs[1, 1]

# Set axis limits
ax_block.set_xlim(min(x)-1, max(x)+1)
ax_block.set_ylim(-1, 1)
ax_block.set_title("Block motion")
ax_block.set_yticks([])
ax_block.axhline(y=-0.06, color='black', linewidth=5)

ax_pos.set_xlim(0, t_stop)
ax_pos.set_ylim(min(x)-1, max(x)+1)
ax_pos.set_title("Position vs Time")

ax_ekin.set_xlim(0, t_stop)
ax_ekin.set_ylim(0, max(E_kin)*1.1)
ax_ekin.set_title("Kinetic Energy vs Time")

ax_work.set_xlim(0, t_stop)
ax_work.set_ylim(min(W)*1.1, max(W)*1.1)
ax_work.set_title("Work vs Time")

# Artists
block_dot, = ax_block.plot([], [], 'rs', markersize=10)
line_pos, = ax_pos.plot([], [], 'b')
line_ekin, = ax_ekin.plot([], [], 'g')
line_work, = ax_work.plot([], [], 'r')

#  Init 
def init():
    block_dot.set_data([], [])
    line_pos.set_data([], [])
    line_ekin.set_data([], [])
    line_work.set_data([], [])
    return block_dot, line_pos, line_ekin, line_work

#  Update function 
def update(frame):
    t = t_values[:frame]
    block_dot.set_data([x[frame]], [0])
    line_pos.set_data(t, x[:frame])
    line_ekin.set_data(t, E_kin[:frame])
    line_work.set_data(t, W[:frame])
    return block_dot, line_pos, line_ekin, line_work

#  Animation 
ani = FuncAnimation(fig, update, frames=len(t_values),
                    init_func=init, interval=dt*1000, blit=True)

plt.close(fig)  # Prevent double static plot
HTML(ani.to_jshtml())

3.8.4Answers set 2¶

Solution to Exercise 9

$W=\Delta E_{kin} = \int_0^x F \mathrm{d}x = \int_0^x k x dx = 1/2kx^2 = 1/2mv^2 \Rightarrow v = \sqrt{\frac{kx^2}{m}}$
The spring has mass as well.
The gravitational does work as well ( $W=F_g\mathrm{d}x \lt 0$ )

Solution to Exercise 13

Solution to Exercise 15

Work done by electric field when the electron moves from $x=\frac{1}{4}L$ to $x=0$ :

W=\int_{\frac{1}{4}L}^{0}{\vec{F}\ .d\vec{s}}=-eE_0\int_{\frac{1}{4}L}^{0}{\sin\left(2\pi\frac{x}{L}\right)dx=\\ -eE_0\frac{L}{2\pi}}\left[-\cos\left(2\pi\frac{x}{L}\right)\right]_{\frac{1}{4}L}^0=\frac{1}{2\pi}eE_0L

(6)

Work done is gain in kinetic energy: $\Delta E_{kin}=W$ . Assuming the only work done is by the electric field and using initial velocity is zero: $v_i=0$ :

\frac{1}{2}mv^2=\ \frac{1}{2\pi}eE_0L\Rightarrow v=\ \sqrt{\frac{eE_0L}{\pi m}}

(7)

Note that indeed the work done is positive, as it should in this case since the electron starts with zero velocity.

Solution to Exercise 16

W=\int_{0}^{L}{\vec{F}\ .d\vec{s}}=\int_{0}^{L}{F_0e^{-\frac{t}{\tau}}\ dx}

(8)

Particle velocity is $v_0=const.$ Thus, trajectory $x\left(t\right)=v_0t$ since at $t=0\rightarrow x=0$ Consequently: $x=L\rightarrow t=\ \frac{L}{v_0}$

Thus, we can write for the amount of work done:

W=\int_{0}^{\frac{L}{v_0}}{F_0e^{-\frac{t}{\tau}}\cdot v_0dt}=\\ F_0v_0\left[-\tau e^{-\frac{t}{\tau}}\right]_0^{L/v_0}=F_0v_0\tau\left(1-e^{-\frac{L}{v_0\tau}}\right)

(9)

We note: $W>0$ and naively, we could expect that the kinetic energy of the particle would have increased. But that isn’t the case: it started with $E_{kin}= \frac{1}{2}mv_0^2$ and it kept this along the entire path as it is given that the particle is traveling with a constant velocity.

From this last statement, we immediately learn, that there must be a second force acting on the particle. This force is exactly equal and opposite to $F$ at all times! Otherwise, the particle would accelerate and change its velocity. Consequently, this second force also perform work on $m$ , the amount is exactly $-W$ and thus the total work done on the particle is zero which reflects that the particle does not change its kinetic energy.

3.8Examples, exercises and solutions

3.8.1Exercise set 1¶

3.8.2Answers set 1¶

3.8.3Exercise set 2¶

3.8.4Answers set 2¶