| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ R = \frac{d}{2}=\frac{30\,\text{cm}}{2}=15\,\text{cm}=0.15\,\text{m} \] | The geometric radius \(R\) of the disk is half its diameter. |
| 2 | \[ a = \frac{R}{2}=\frac{0.15\,\text{m}}{2}=0.075\,\text{m} \] | The axle (black dot) is halfway from the centre to the rim, so its horizontal displacement from the centre is \(a\). |
| 3 | \[ \begin{aligned} \ell_{\text{L}} &= R-a = 0.15-0.075 = 0.075\,\text{m}\\[4pt] \ell_{\text{R}} &= R+a = 0.15+0.075 = 0.225\,\text{m}\\[4pt] \ell_{\text{disk}} &= a = 0.075\,\text{m} \end{aligned} \] | With the axle as the pivot, the perpendicular (horizontal) distances to each vertical force are: • left rope: \(R − a\) • right rope: \(R + a\) • disk weight (acting at the centre): \(a\). |
| 4 | \[ \begin{aligned} \tau_{\text{L}} &= (+)\,m_{\text{L}} g\, \ell_{\text{L}} = (15\,\text{kg})(9.8\,\text{m/s}^2)(0.075\,\text{m}) = +11.025\,\text{N·m}\\[4pt] \tau_{\text{R}} &= (-)\,m_{\text{R}} g\, \ell_{\text{R}} = (10\,\text{kg})(9.8\,\text{m/s}^2)(0.225\,\text{m}) = -22.05\,\text{N·m}\\[4pt] \tau_{\text{disk}} &= (-)\,m_{\text{disk}} g\, \ell_{\text{disk}} = (5.0\,\text{kg})(9.8\,\text{m/s}^2)(0.075\,\text{m}) = -3.675\,\text{N·m} \end{aligned} \] | Take counter-clockwise (CCW) as positive. • The left hanging mass tends to rotate the disk CCW. • The right hanging mass and the disk’s own weight both tend to rotate it clockwise (negative). |
| 5 | \[ \tau_{\text{ext}} = \tau_{\text{L}} + \tau_{\text{R}} + \tau_{\text{disk}} = 11.025 – 22.05 – 3.675 = -14.70\,\text{N·m} \] | Sum of all external torques about the axle (excluding the axle itself). |
| 6 | \[ \tau_{\text{axle}} + \tau_{\text{ext}} = 0 \;\;\Rightarrow\;\; \boxed{\tau_{\text{axle}} = +14.7\,\text{N·m}} \] | For rotational equilibrium the net torque must vanish; therefore the axle must supply an equal and opposite torque. The positive sign indicates a counter-clockwise torque supplied by the axle. |
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A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
The system in the Figure is in equilibrium. A concrete block of mass \(225 \, \text{kg}\) hangs from the end of a uniform strut whose mass is \(45.0 \, \text{kg}\).
A child of mass \( 3 \) \( \text{kg} \) rotates on a platform of \( 10 \) \( \text{kg} \). They start walking towards the center while the platform is rotating. Which of the following could possibly decrease the total angular momentum of the child-platform system?
A meterstick is supported at its center, which is aligned with the center of a cradle located at position \( x = 0 \) \( \text{m} \). Two identical objects of mass \( 1.0 \) \( \text{kg} \) are suspended from the meterstick. One object hangs \( 0.25 \) \( \text{m} \) to the left of the support point, and the other object hangs \( 0.50 \) \( \text{m} \) to the right of the support point. The system is released from rest and is free to rotate. Which of the following claims correctly describes the subsequent motion of the system containing the meterstick, cradle, and the two objects?
A student is asked to design an experiment to determine the change in angular momentum of a disk that rotates about its center and the product of the average torque applied to the disk and the time interval in which the torque is exerted. A net force is applied tangentially to the surface of the disk. The rotational inertia of the disk about its center is \(I = MR^2\). Which two of the following quantities should the student measure to determine the change in angular momentum of the disk after 10 s? Select two answers.
A rod of length \( L \) is rotated about its center with \( I = \frac{ML^{2}}{12} \). What is the moment of inertia at either end of the rod?
Initially, a ball has an angular velocity of \( 5.0 \) \( \text{rad/s} \) counterclockwise. Some time later, after rotating through a total angle of \( 5.5 \) \( \text{radians} \), the ball has an angular velocity of \( 1.5 \) \( \text{rad/s} \) clockwise.
| Wagon | Wheel Structure | Moment of Inertia | Wheel Mass | Wheel Radius |
|---|---|---|---|---|
| Wagon \(A\) | Solid disk | \[\frac{1}{2} M R^2\] | \[ 0.5 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
| Wagon \(B\) | Solid disk | \[\frac{1}{2} M R^2\] | \[ 0.2 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
| Wagon \(C\) | Hollow hoop | \[M R^2\] | \[ 0.1 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
Three wagons have identical total mass (including their wheels) and each has four wheels. However, the wheels on each wagon have different designs with varying mass distributions and radii as shown in a reference chart. When accelerating each wagon from a standstill to \( 10 \) \( \text{m/s} \), which wagon requires the most energy input?
A uniform rod of length \( L \) is pivoted at one end \(45^{\circ}\) below the horizontal and released from rest. The rod swings freely downward. Which of the following best describes the angular acceleration of the rod as it swings from the initial position to the vertical position?
A solid ball of mass \( M \) and radius \( R \) has rotational inertia \( \frac{2}{5} M R^{2} \) about its center. It rolls without slipping along a level surface at speed \( v \) just before it begins rolling up an inclined plane. Which of the following expressions correctly represents the maximum vertical height the solid ball can ascend to when it rolls up the incline without slipping?
\(+14.7\,\text{N·m}\)
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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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