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 block hangs from the ceiling by a massless rope. A \( 3.0 \, \text{kg} \) block is attached to the first block and hangs below it on another piece of massless rope. The tension in the top rope is \( 63.0 \, \text{N} \).
The figure shows a truck pulling three crates across a rough road. Which of the following best describes the directions of all the horizontal forces acting on crate 2?
Two masses, \( m_1 \) and \( m_2 \), are connected by a cord and arranged as shown in the diagram, with \( m_1 \) sliding along a frictionless surface and \( m_2 \) hanging from a light, frictionless pulley. What would be the mass of the falling mass, \( m_2 \), if both the sliding mass, \( m_1 \), and the tension, \( T \), in the cord were known?
A net force of \( 8.0 \) \( \text{N} \) accelerates a \( 4.0 \) \( \text{kg} \) body from rest to a speed of \( 5.0 \) \( \text{m s}^{-1} \). Which of the following is equal to the work done by the force?
A \( 25.0 \) \( \text{kg} \) block is placed at the top of an inclined plane set at an angle of \( 35 \) degrees to the horizontal. The block slides down the \( 1.5 \) \( \text{m} \) slope at a constant rate. How much work did friction do on the block?
Two wires support an unknown mass as shown in the diagram. The tension in the left wire is measured to be \( 17.5 \) \( \text{N} \) and the tension in the right wire is \( 30.3 \) \( \text{N} \). The left wire makes an angle of \( 30^{\circ} \) with the horizontal, and the right wire makes an angle of \( 60^{\circ} \) with the horizontal. What is the mass of the object?
A person is running on a track. Which of the following forces propels the runner forward?
A block is initially at rest on top of an inclined ramp that makes an angle \( \theta_0 \) with the horizontal. The distance measured along the base of the ramp is \( D \). After the block is released from rest, it slides down the frictionless ramp and then continues onto a rough horizontal surface until it finally comes to rest at the position \( x = 4D \) measured from the base of the ramp. The coefficient of kinetic friction between the block and the rough horizontal surface is \( \mu_k \).
A car travels to right at constant velocity. The net force on the car is
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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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