| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T = 2\pi \sqrt{\frac{I}{mg\,\Delta x}}\] | This is the period formula for a physical pendulum, where \(I\) is the moment of inertia and \(\Delta x\) is the distance from the pivot to the center of mass. |
| 2 | \[\Delta x = r\] | Since the ornament (a thin hollow sphere) is suspended by a wire attached at its top, the distance from the pivot to its center of mass is the radius \(r\) of the sphere. |
| 3 | \[T = 2\pi \sqrt{\frac{I}{mgr}}\] | Substitute \(\Delta x = r\) into the period formula. |
| 4 | \[I = \frac{5}{3}mr^2\] | The moment of inertia about the pivot is given. But you can also use the parallel axis theorem for a thin hollow sphere: \(I = I_{\text{cm}} + mr^2 = \frac{2}{3}mr^2 + mr^2 = \frac{5}{3}mr^2\). |
| 5 | \[T = 2\pi \sqrt{\frac{\frac{5}{3}mr^2}{mgr}} = 2\pi \sqrt{\frac{5r}{3g}}\] | Substitute \(I = \frac{5}{3}mr^2\) into the period expression and cancel the common factors \(m\) and \(r\). |
| 6 | \[T = \frac{1}{f}\] | Relate the period \(T\) with the frequency \(f\) using \(f = 1/T\). Given \(f = 2.50\;\text{Hz}\), \(T = \frac{1}{2.50}\). |
| 7 | \[\frac{1}{f} = 2\pi \sqrt{\frac{5r}{3g}}\] | Express the period in terms of the frequency and set it equal to the derived expression from step 5. |
| 8 | \[\left(\frac{1}{f}\right)^2 = 4\pi^2 \frac{5r}{3g}\] | Square both sides to eliminate the square root. |
| 9 | \[\frac{5r}{3g} = \frac{1}{4\pi^2 f^2}\] | Solve for the expression containing \(r\) by isolating it on one side. |
| 10 | \[r = \frac{3g}{5} \cdot \frac{1}{4\pi^2 f^2} = \frac{3g}{20\pi^2 f^2}\] | Solve for \(r\) by multiplying both sides appropriately. |
| 11 | \[r = \frac{3(9.80)}{20\pi^2 (2.50)^2}\] | Substitute \(g = 9.80\;\text{m/s}^2\) and \(f = 2.50\;\text{Hz}\) into the expression. |
| 12 | \[r \approx \frac{29.4}{1233.7} \approx 0.024\;\text{m}\] | Calculate the numerical value |
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A simple pendulum oscillates with amplitude \(A\) and period \(T\), as represented on the graph above. Which option best represents the magnitude of the pendulum’s velocity \(v\) and acceleration \(a\) at time \(\frac{T}{2}\)?
An experimenter has a simple pendulum of length \( L \) and a mass–spring system with mass \( m \) and spring constant \( k \). Both are found to have the same period of oscillation \( T \) on Earth. If both systems are taken to the Moon, where the acceleration due to gravity is approximately \( \frac{1}{6} g \) of Earth, what will happen to their periods?
Two uniform disks have the same mass but different radii: disk \( 1 \) has a radius \( R \) and disk \( 2 \) has a radius \( 2R \). What is the ratio of the moment of inertia of the first disk to the second disk?
The object shown in the diagram below consists of a cylinder of mass \( 100 \) \( \text{kg} \) and radius \( 25.0 \) \( \text{cm} \) connected by four thin rods, each of mass \( 5.00 \) \( \text{kg} \) and length \( 0.75 \) \( \text{m} \), to a thin-outer ring of mass \( 20.0 \) \( \text{kg} \). A small chunk of metal of mass \( 1.00 \) \( \text{kg} \) is welded to the outer ring. Determine the moment of inertia of the entire assembly about the center of the inner cylinder, treating the metal chunk as a point mass. Hint: The moment of inertia of a disk about it center is \(\tfrac{1}{2} M R^2\), a thin rod about it center is \(\tfrac{1}{12}ML^2\), and a thin hoop about its center is \(I = MR^2\).
What is the effect on the period of a pendulum if you double its length?
A string is wound tightly around a fixed pulley having a radius of 5.0 cm. As the string is pulled, the pulley rotates without any slipping of the string. What is the angular speed of the pulley when the string is moving at 5.0 m/s?
A sphere starts from rest and rolls down an incline of height \( H = 1.0 \) \( \text{m} \) at an angle of \( 25^\circ \) with the horizontal, as shown above. The radius of the sphere \( R = 15 \) \( \text{cm} \), and its mass \( m = 1.0 \) \( \text{kg} \). The moment of inertia for a sphere is \( \frac{2}{5}mR^2 \). What is the speed of the sphere when it reaches the bottom of the plane?
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?
What is the relationship between the period \( T \) and frequency \( f \) of an object in simple harmonic motion?
Three pendulums are set in motion, oscillating through small amplitudes. Each has the same mass. Rank the period of the pendulums from shortest to longest.
\(\boxed{r \approx 0.024\;\text{m}}\)
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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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