| 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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An object moves at a constant speed of \( 9.0 \frac{m}{s} \) in a circular path of radius of 1.5 m. What is the angular acceleration of the object?
A uniform, solid, \( 100 \) \( \text{kg} \) cylinder with a diameter of \( 1.0 \) \( \text{m} \) is mounted so it is free to rotate about a fixed, horizontal, frictionless axis that passes through the centers of its circular ends. A \( 10 \) \( \text{kg} \) block is hung from a very light, thin cord wrapped around the cylinder’s circumference. When the block is released, the cord unwinds and the block accelerates downward. What is the acceleration of the block?
A solid sphere \( \left( I = \frac{2}{5}MR^2 \right) \) and a solid cylinder \( \left( I = \frac{1}{2}MR^2 \right) \), both uniform and of the same mass and radius, roll without slipping at the same forward speed. It is correct to say that the total kinetic energy of the solid sphere is
Wheels \( A \) and \( B \) are connected by a moving belt and are both free to rotate about their centers. The belt does not slip on the wheels. The radius of Wheel \( B \) is twice the radius of Wheel \( A \). Wheel \( A \) has constant angular speed \( \omega_A \) and Wheel \( B \) has constant angular speed \( \omega_B \). Which of the following correctly relates \( \omega_A \) and \( \omega_B \)?
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?
Consider a rigid body that is rotating. Which of the following is an accurate statement?
A hoop with a mass \(m\) and unknown radius is rolling without slipping on a flat surface with an angular speed \(\omega\). The hoop encounters a hill and continues to roll without slipping until it reaches a maximum height \(h\).
A \( 6.00 \, \text{m} \) long, \( 500 \, \text{kg} \) steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A \( 70 \, \text{kg} \) construction worker stands at the far end of the beam. What is the magnitude of the torque about the bolt due to the worker and the weight of the beam?
To increase the moment of inertia of a body about an axis, you must
\(+14.7\,\text{N·m}\)
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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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