| Derivation or Formula | Reasoning |
|---|---|
| \[m = 25\,\text{g} = 0.025\,\text{kg},\quad d = 24\,\text{cm} = 0.24\,\text{m},\quad r = \frac{d}{2} = 0.12\,\text{m}\] | Convert to SI units and compute the wheel radius \(r\) from the given diameter \(d\). |
| \[\alpha = 470\,\text{rad/s}^2,\quad \Delta\theta = \frac{3}{4}(2\pi) = \frac{3\pi}{2}\] | The angular acceleration is given, and \(\tfrac{3}{4}\) of a revolution corresponds to angular displacement \(\Delta\theta\). |
| \[\tau = I\alpha\] | Torque \(\tau\) relates to angular acceleration \(\alpha\) through rotational dynamics. |
| \[I = mr^2\] | The wheel is negligible mass, so the rotating inertia is dominated by the ball modeled as a point mass at radius \(r\). |
| \[W = \tau\Delta\theta\] | Work done by a constant torque over an angular displacement is \(W = \tau\Delta\theta\). |
| \[W = (I\alpha)\Delta\theta = (mr^2\alpha)\Delta\theta\] | Substitute \(\tau = I\alpha\) and \(I = mr^2\) into the work expression, as indicated in the hint. |
| \[mr^2\alpha\Delta\theta = mgh\] | Use conservation of energy from the very start (rest) to the maximum height: the work put in becomes gravitational potential energy \(mgh\). |
| \[r^2\alpha\Delta\theta = gh\] | Cancel \(m\) from both sides, showing the height does not depend on the ball’s mass. |
| \[h = \frac{r^2\alpha\Delta\theta}{g}\] | Solve algebraically for the maximum height above the starting reference implied by the energy equation. |
| \[h = \frac{(0.12)^2(470)\left(\frac{3\pi}{2}\right)}{9.8}\] | Substitute \(r = 0.12\,\text{m}\), \(\alpha = 470\,\text{rad/s}^2\), \(\Delta\theta = \frac{3\pi}{2}\), and \(g = 9.8\,\text{m/s}^2\). |
| \[h = \frac{(0.0144)(470)\left(\frac{3\pi}{2}\right)}{9.8} = \frac{6.768\left(\frac{3\pi}{2}\right)}{9.8}\] | Compute \(r^2\) and multiply by \(\alpha\) to simplify the numerator step-by-step using algebra. |
| \[h = \frac{6.768(1.5\pi)}{9.8} = \frac{10.152\pi}{9.8}\] | Rewrite \(\frac{3\pi}{2} = 1.5\pi\) and multiply constants. |
| \[\boxed{h \approx 3.25\,\text{m above the center}}\] | The maximum height reached above the center of the wheel. |
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A pendulum consists of a ball of mass \( m \) suspended at the end of a massless cord of length \( L \). The pendulum is drawn aside through an angle of \( 60^\circ \) with the vertical and released. At the low point of its swing, the speed of the pendulum ball is
List at least 2 everyday forces that are not conservative, and explain why they aren’t.
A box of mass \( 20 \) \( \text{kg} \) moves to the right on a horizontal frictionless surface with a speed of \( 4.0 \) \( \text{m/s} \). The box collides with and remains attached to one end of a spring of negligible mass whose other end is fixed to a wall. After the collision, the spring compresses a maximum distance of \( 0.50 \) \( \text{m} \), and the box then oscillates back and forth.
The two blocks of masses \( M \) and \( 2M \) travel at the same speed \( v \) but in opposite directions. They collide and stick together. How much mechanical energy is lost to other forms of energy during the collision?
| Speed | \( 10 \, \mathrm{m/s} \) | \( 20 \, \mathrm{m/s} \) | \( 30 \, \mathrm{m/s} \) |
| Braking Distance | \( 6.1 \, \mathrm{m} \) | \( 23.9 \, \mathrm{m} \) | \( 53.5 \, \mathrm{m} \) |
A car of mass \( 1500 \, \mathrm{kg} \) is traveling at one of the speeds listed when the brakes are first applied. Using the data above, what is the magnitude of the average braking force required to stop the car?
A uniform stick has length \( L \). The moment of inertia about the center of the stick is \( I_0 \). A particle of mass \( M \) is attached to one end of the stick. The moment of inertia of the combined system about the center of the stick is
How long does it take for a rotating object to speed up from 15.0 rad/s to 33.3 rad/s if it has a uniform angular acceleration of 3.45 rad/s2?
A \( 0.20 \) \( \text{kg} \) object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown in the graph above. The object starts from rest at displacement \( x = 0 \) and time \( t = 0 \) and is displaced a distance of \( 20 \) \( \text{m} \). Determine each of the following.
A stone is falling at a constant velocity vertically down a tube filled with oil. Which of the following statements about the energy changes of the stone during its motion are correct?
I. The gain in kinetic energy is less than the loss in gravitational potential energy.
II. The sum of kinetic and gravitational potential energy of the stone is constant.
III. The work done by the force of gravity has the same magnitude as the work done by friction.
A snowboarder starts from rest and slides down a \(32^\circ\) incline that’s \(75 \, \text{m}\) long.
\(3.25\,\text{m}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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