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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ K_{\text{trans}} = \frac{1}{2} M_{\text{tot}} v^2 \] | All three wagons have the same total mass and are accelerated to the same speed, so the translational kinetic energy is identical for each. |
| 2 | \[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \quad \text{with} \quad \omega = \frac{v}{R} \] | This is the rotational kinetic energy for a wheel rolling without slipping. |
| 3 | \[ I_{\text{disk}} = \frac{1}{2} M_{w} R^2 \quad \Rightarrow \quad K_{\text{rot,disk}} = \frac{1}{2}\left(\frac{1}{2} M_{w} R^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{4} M_{w} v^2 \] | For a solid disk wheel (used in Wagons A and B), the moment of inertia is substituted to find its rotational kinetic energy. |
| 4 | \[ I_{\text{hoop}} = M_{w} R^2 \quad \Rightarrow \quad K_{\text{rot,hoop}} = \frac{1}{2}\left(M_{w} R^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{2} M_{w} v^2 \] | For a hollow hoop wheel (used in Wagon C), the moment of inertia is larger, leading to greater rotational energy. |
| 5 | \[ K_{\text{total, wheel}} = K_{\text{trans, wheel}} + K_{\text{rot}} = \frac{1}{2} M_{w} v^2 + K_{\text{rot}} \] | Each wheel has translational kinetic energy (as it moves with the wagon) plus its rotational kinetic energy. |
| 6 | For a disk: \[ K_{\text{total, disk}} = \frac{1}{2} M_{w} v^2 + \frac{1}{4} M_{w} v^2 = \frac{3}{4} M_{w} v^2 \] |
This is the effective energy per disk wheel. If the wheel’s mass were a simple point mass, it would only have \(\frac{1}{2}M_{w}v^2\); thus, the extra energy due to rotation is \(\Delta K_{\text{disk}} = \frac{3}{4} M_{w} v^2 – \frac{1}{2} M_{w} v^2 = \frac{1}{4} M_{w} v^2\). |
| 7 | For a hoop: \[ K_{\text{total, hoop}} = \frac{1}{2} M_{w} v^2 + \frac{1}{2} M_{w} v^2 = M_{w} v^2 \] |
Similarly, the extra energy due to rotation for a hoop is \(\Delta K_{\text{hoop}} = M_{w} v^2 – \frac{1}{2} M_{w} v^2 = \frac{1}{2} M_{w} v^2\), which is larger than that for a disk of the same mass. |
| 8 | Wagon A (disk, \(M_{w}=0.5\,\text{kg}\)): \[ \Delta K_{A} = 4\left(\frac{1}{4}\times 0.5\, v^2\right) = 4(0.125\, v^2) = 0.5\, v^2 \]Wagon B (disk, \(M_{w}=0.2\,\text{kg}\)): \[ \Delta K_{B} = 4\left(\frac{1}{4}\times 0.2\, v^2\right) = 4(0.05\, v^2) = 0.2\, v^2 \]Wagon C (hoop, \(M_{w}=0.2\,\text{kg}\)): \[ \Delta K_{C} = 4\left(\frac{1}{2}\times 0.2\, v^2\right) = 4(0.1\, v^2) = 0.4\, v^2 \] |
Each wagon has four wheels. The extra energy due to wheel rotation is calculated by multiplying the additional energy per wheel by four. |
| 9 | With \(v = 10\,\text{m/s}\) (\(v^2 = 100\)): \[ \Delta K_{A} = 0.5 \times 100 = 50\, \text{J} \] \[ \Delta K_{B} = 0.2 \times 100 = 20\, \text{J} \] \[ \Delta K_{C} = 0.4 \times 100 = 40\, \text{J} \] |
Substitute the final speed into each expression to obtain the numerical extra energy that must be supplied to accelerate the wheels. |
| 10 | \[\boxed{\text{Wagon A}}\] | Since the translational energy is the same for all wagons, the wagon with the greatest extra wheel energy (50 J) requires the most energy input. Therefore, Wagon A requires the most energy. |
Just ask: "Help me solve this problem."
The moment of inertia of a solid cylinder about its axis is given by \( I = \frac{1}{2}mR^2 \). If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
A disk of radius \( R = 0.5 \) \( \text{cm} \) rests on a flat, horizontal surface such that frictional forces are considered to be negligible. Three forces of unknown magnitude are exerted on the edge of the disk, as shown in the figure. Which of the following lists the essential measuring devices that, when used together, are needed to determine the change in angular momentum of the disk after a known time of \( 5.0 \) \( \text{s} \)?
A meter stick of mass [katex] .2 [/katex] kg is pivoted at one end and supported horizontally. A force of [katex] 3 [/katex] N downwards is applied to the free end, perpendicular to the length of the meter stick. What is the net torque about the pivot point?
Why are doorknobs located on the side of the door opposite the hinges?
The elliptical orbit of a comet is shown above. Positions 1 and 2 are, respectively, the farthest and nearest positions to the Sun, and at position 1 the distance from the comet to the Sun is 10 times that at position 2. What is the ratio \(v_1\)/\(v_2\) of the speed of the comet at position 1 to the speed at position 2?
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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] |
General Metric Conversion Chart
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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