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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."
A horizontal uniform rod of length L and mass M is pivoted at one end and is initially at rest. A small ball of mass M (same masses) is attached to the other end of the rod. The system is released from rest. What is the angular acceleration of the rod just immediately after the system is released?
The moment of inertia of a solid cylinder about its axis is given by \( 0.5MR^2 \). If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
Find the following three values using just rotational kinematics.
Consider a rigid body that is rotating. Which of the following is an accurate statement?
A uniform ladder with mass \( m_2 \) and length \( L \) rests against a smooth wall. A do-it-yourself enthusiast of mass \( m_1 \) stands on the ladder a distance \( d \) from the bottom (measured along the ladder). The ladder makes an angle \( \theta \) with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude \( f \) between the floor and the ladder. \( N_1 \) is the magnitude of the normal force exerted by the wall on the ladder, and \( N_2 \) is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive.
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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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