| 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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One end of a string is wrapped around a pulley that is free to rotate with negligible friction about an axle at its center. The other end of the string is attached to a block. The block is released from rest and moves downward with constant acceleration. Which of the following correctly indicates whether the amount of work done on the pulley by the string during each successive complete rotation remains constant or increases, and provides a valid justification?
Why are doorknobs located on the side of the door opposite the hinges?
Three forces of equal magnitude are applied to a \(3 \, \text{m} \times 2 \, \text{m}\) rectangle. Force \(F_1\) and \(F_2\) act at \(45^\circ\) angles to the vertical as shown, while \(F_3\) acts horizontally.
In short:
\(F_1\): applied at \((0, -2)\), direction SW \(45^\circ\)
\(F_2\): applied at \((2, -2)\), direction NW \(45^\circ\)
\(F_3\): applied at \((3, -1)\), direction east
Points of rotation: \(A = (0, 0)\), \(B = (0, -1)\), \(C = (1, -1)\)
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?
The driver of a car traveling at \( 30.0 \) \( \text{m/s} \) applies the brakes and undergoes a constant negative acceleration of \( 2.00 \) \( \text{m/s}^2 \). How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of \( 0.300 \) \( \text{m} \)?
Find the following three values using just rotational kinematics.
Which of the following must be true for an object at translational equilibrium?
A rod may freely rotate about an axis that is perpendicular to the rod and is along the plane of the page. The rod is divided into four sections of equal length of 0.2 m each, and four forces are exerted on the rod, as shown in the figure. Frictional forces are considered negligible. Which of the following describes an additional torque that must be applied in order to keep the rod from rotating?
The downward motion of an elevator is controlled by a cable that unwinds from a cylinder of radius \( 0.20 \) \( \text{m} \). What is the angular velocity of the cylinder when the downward speed of the elevator is \( 1.2 \) \( \text{m/s} \)?
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?
\(+14.7\,\text{N·m}\)
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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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