| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta L_A = 10\,\text{kg}\cdot\text{m}^2/\text{s}\] | The graph shows disk A rises from \(0\) to \(10\,\text{kg}\cdot\text{m}^2/\text{s}\) between \(t = 0\,\text{s}\) and \(t = 5\,\text{s}\), so \(\Delta L_A = 10\). |
| 2 | \[\Delta L_B = 20\,\text{kg}\cdot\text{m}^2/\text{s}\] | The graph shows disk B rises from \(0\) to \(20\,\text{kg}\cdot\text{m}^2/\text{s}\) in the same time interval, so \(\Delta L_B = 20\). |
| 3 | \[\tau_{\text{avg}} = \frac{\Delta L}{\Delta t}\] | Average net torque equals the change in angular momentum divided by the time interval (algebraic definition, no calculus needed). |
| 4 | \[\tau_A = \frac{10}{5} = 2\,\text{N}\cdot\text{m}\] | Substituting \(\Delta L_A = 10\) and \(\Delta t = 5\,\text{s}\) gives the average torque on disk A. |
| 5 | \[\tau_B = \frac{20}{5} = 4\,\text{N}\cdot\text{m}\] | Substituting \(\Delta L_B = 20\) and \(\Delta t = 5\,\text{s}\) gives the average torque on disk B. |
| 6 | \[\tau_B = 2\tau_A\] | Using \(\tau_A = 2\) and \(\tau_B = 4\) shows the required relationship. The disks’ different rotational inertias are irrelevant because inertia cancels in \(\Delta L\). |
| Incorrect Option (a) | \[\tau_B = 4\tau_A\] | This predicts \(\tau_B = 8\,\text{N}\cdot\text{m}\), which contradicts the calculated \(\tau_B = 4\,\text{N}\cdot\text{m}\). |
| Incorrect Option (c) | \[\tau_B = \tfrac{1}{2}\tau_A\] | This would give \(\tau_B = 1\,\text{N}\cdot\text{m}\), far below the value obtained from the graph. |
| Incorrect Option (d) | \[\tau_B = \tfrac{1}{4}\tau_A\] | This would give \(\tau_B = 0.5\,\text{N}\cdot\text{m}\), also inconsistent with the calculated \(4\,\text{N}\cdot\text{m}\). |
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A solid ball and a cylinder roll down an inclined plane. Which reaches the bottom first?
Which of the following situations will increase the moment of inertia of a solid cylinder \( I = \tfrac{1}{2} M R^{2} \) by the same amount?
The rotating systems, shown in the figure above, differ only in that the two identical movable masses are positioned a distance r from the axis of rotation (left), or a distance r/2 from the axis of rotation (right). What happens if you release the hanging blocks simultaneously from rest?
A disk is initially rotating counterclockwise around a fixed axis with angular speed \( \omega_0 \). At time \( t = 0 \), the two forces shown in the figure above are exerted on the disk. If counterclockwise is positive, which of the following could show the angular velocity of the disk as a function of time?
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of \( 5.0 \) \( \text{rev/s} \) in \( 8.0 \) \( \text{s} \). At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in \( 12.0 \) \( \text{s} \). Through how many revolutions does the tub turn during the entire \( 20 \)-s interval? Assume constant angular acceleration while it is starting and stopping.
A uniform, rigid rod of length \( 2 \) \( \text{m} \) lies on a horizontal surface. One end of the rod can pivot about an axis that is perpendicular to the rod and along the plane of the page. A \( 10 \) \( \text{N} \) force is applied to the rod at its midpoint at an angle of \( 37^{\circ} \). A second force \( F \) is applied to the free end of the rod so that the rod remains at rest, as shown in the figure. The magnitude of the torque produced by force \( F \) is most nearly
The system above is NOT balanced since \(m_2\) is twice the mass of \(m_1\). Which of the following changes would NOT balance the system so that there is 0 net torque? Assume the plank has no mass of its own.
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.
Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the second force is applied at \( 30^\circ \) to the plane of the door. Which force exerts the greater torque about the door hinge?
A meter stick with a uniformly distributed mass of \(0.5 \, \text{kg}\) is supported by a pivot placed at the \(0.25 \, \text{m}\) mark from the left. At the left end, a small object of mass \(1.0 \, \text{kg}\) is placed at the zero mark, and a second small object of mass \(0.5 \, \text{kg}\) is placed at the \(0.5 \, \text{m}\) mark. The meter stick is supported so that it remains horizontal, and then it is released from rest. Find the change in the angular momentum of the meter stick, one second after it is released.
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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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