Part A – Angular Speed
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T\cos(\theta)=mg\] | The vertical component of the tension \(T\) balances the gravitational force \(mg\) on the ball. |
| 2 | \[T\sin(\theta)=m\omega^2r\] | The horizontal component of \(T\) provides the centripetal force (\(m\omega^2r\)) needed for circular motion. |
| 3 | \[\frac{T\sin(\theta)}{T\cos(\theta)}=\frac{m\omega^2r}{mg}\] | Dividing the horizontal equation by the vertical one eliminates \(T\) to relate \(\omega\) and \(\theta\). |
| 4 | \[\tan(\theta)=\frac{\omega^2r}{g}\] | This simplifies the relationship between the angle \(\theta\) and the angular speed \(\omega\). |
| 5 | \[\omega^2=\frac{g\tan(\theta)}{r}\] | Solving for \(\omega^2\) in terms of \(\tan(\theta)\), \(r\), and \(g\). |
| 6 | \[\sin(\theta)=\frac{r}{\ell}\quad \text{and}\quad \cos(\theta)=\sqrt{1-\frac{r^2}{\ell^2}}=\frac{\sqrt{\ell^2-r^2}}{\ell}\] | Using the geometry of the conical pendulum, where the horizontal radius \(r\) relates to the string length \(\ell\) and angle \(\theta\). |
| 7 | \[\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{r}{\sqrt{\ell^2-r^2}}\] | Expressing \(\tan(\theta)\) in terms of the given variables \(r\) and \(\ell\). |
| 8 | \[\omega^2=\frac{g}{r}\cdot\frac{r}{\sqrt{\ell^2-r^2}}=\frac{g}{\sqrt{\ell^2-r^2}}\] | Substituting the expression for \(\tan(\theta)\) into the equation for \(\omega^2\) simplifies the result. |
| 9 | \[\boxed{\omega=\sqrt{\frac{g}{\sqrt{\ell^2-r^2}}}}\] | Taking the square root yields the final expression for the angular speed \(\omega\) in terms of \(\ell\), \(r\), and \(g\). |
Part B – Tension
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T\cos(\theta)=mg\] | The vertical component of the tension balances the weight of the ball. |
| 2 | \[T=\frac{mg}{\cos(\theta)}\] | Solving for the tension \(T\) from the vertical equilibrium equation. |
| 3 | \[\cos(\theta)=\frac{\sqrt{\ell^2-r^2}}{\ell}\] | Expressing \(\cos(\theta)\) in terms of the string length \(\ell\) and the horizontal radius \(r\) using geometry. |
| 4 | \[T=\frac{mg}{\frac{\sqrt{\ell^2-r^2}}{\ell}}=\frac{mg\ell}{\sqrt{\ell^2-r^2}}\] | Substituting \(\cos(\theta)\) into the equation for \(T\) and simplifying. |
| 5 | \[\boxed{T=\frac{mg\ell}{\sqrt{\ell^2-r^2}}}\] | This is the final expression for the tension \(T\) in the string in terms of \(L\) (\(\ell\)), \(m\), \(r\), and \(g\). |
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A person pulls a rope with a force \( F \) at an angle of \( 60^\circ \) to the horizontal. The rope is connected to a load over a frictionless pulley as shown in the diagram. The load is stationary. Which of the following is correct about the weight of the load and the net force exerted on the pulley by the rope?

The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is
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A \( 240 \) \( \text{kg} \) block is dropped from \( 3.0 \) meters onto a spring, compresses the spring and comes to rest.
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Calculate the final speed of the space probe in \( \text{km s}^{-1} \).
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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