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| 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. |
Just ask: "Help me solve this problem."
A rod of length \( L \) is rotated about its center with \( I = \frac{ML^{2}}{12} \). What is the moment of inertia at a point \( \frac{L}{4} \) away from the center?
Two workers are holding a thin plate with length \(5 \, \text{m}\) and height \(2 \, \text{m}\) at rest by supporting the plate in the bottom corners. The workers are standing at rest on a slope of \(10^\circ\). Treat these supporting forces as vertical normal forces and calculate their magnitudes and state if both workers are sharing “the job” fairly.
A system of two wheels fixed to each other is free to rotate about a frictionless axis through the common center of the wheels and perpendicular to the page. Four forces are exerted tangentially to the rims of the wheels, as shown in the figure. The magnitude of the net torque on the system about the axis is
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
Two disks, A and B, each experience a net external torque that varies over an interval of \( 5 \) \( \text{s} \). Disk B has a rotational inertia that is twice that of Disk A. The graph shown represents the angular momentum of the two disks as functions of time between \( t = 0 \) \( \text{s} \) and \( t = 5 \) \( \text{s} \). The average magnitudes of the net torques exerted on disks A and B from \( t = 0 \) \( \text{s} \) to \( t = 5 \) \( \text{s} \) are \( \tau_A \) and \( \tau_B \), respectively. Which of the following expressions correctly relates the magnitudes of the average torques?
\(+14.7\,\text{N·m}\)
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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] |
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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