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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]\tau = r \times F[/katex] | Torque ([katex]\tau[/katex]) is calculated as the product of the radial distance ([katex]r[/katex]) and the force applied ([katex]F[/katex]). The force here could be the result of the weights due to the masses. |
| 2 | [katex]F = mg[/katex] | Force due to gravity is calculated by multiplying the mass ([katex]m[/katex]) by the acceleration due to gravity ([katex]g[/katex], approximately [katex]9.8 \, \text{m/s}^2[/katex]). |
| 3 | [katex] \tau_{\text{left}} = 0.5 \, \text{m} \times 0.1 \, \text{kg} \times 9.8 \, \text{m/s}^2 [/katex] [katex] \tau_{\text{left}} = 0.49 \, \text{Nm} [/katex] |
The torque produced by the mass on the left end. The radial distance here is half a meter since it’s at the end of half the length of the meter stick. |
| 4 | [katex] \tau_{\text{right}} = 0.5 \, \text{m} \times 0.15 \, \text{kg} \times 9.8 \, \text{m/s}^2 [/katex] [katex] \tau_{\text{right}} = 0.735 \, \text{Nm} [/katex] |
The torque produced by the mass on the right end, using similar calculations as for the left side. |
| 5 | [katex] \text{Net } \tau = \tau_{\text{right}} – \tau_{\text{left}} [/katex] [katex] \text{Net } \tau = 0.735 \, \text{Nm} – 0.49 \, \text{Nm} [/katex] [katex] \text{Net } \tau = 0.245 \, \text{Nm} [/katex] |
Calculate the net torque on the meter stick. Since the stick is in rotational equilibrium, this net torque has to be counteracted by the tension in the string. |
| 6 | [katex] T \times 0.5 \, \text{m} = 0.245 \, \text{Nm} [/katex] | Set the torque due to the tension ([katex]T[/katex]) equal to the net torque. The distance from the pivot to the left end is [katex]0.5 \, \text{m}[/katex]. |
| 7 | [katex] T = \frac{0.245 \, \text{Nm}}{0.5 \, \text{m}} [/katex] [katex] T = 0.49 \, \text{N} [/katex] |
Solve for the tension [katex]T[/katex] in the string supporting the left end of the meter stick. |
| 8 | [katex] T = 0.49 \, \text{N} [/katex] | Final answer: The tension in the string supporting the left end of the meter stick is 0.49 Newtons. |
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The angular velocity of an electric motor is \( \omega = \left(20 – \frac{1}{2} t^2 \right) \, \text{rad/s} \), where \(t\) is in seconds.
A car is moving up the side of a circular roller coaster loop of radius 12 m. The angular velocity is [katex] 1.8 \, \text{rad/s} [/katex] and angular acceleration is [katex] -0.82 \, \text{rad/s}^2 [/katex]. The car is at the same elevation as the center of the loop. Find the magnitude and direction of the acceleration.
The coefficient of static friction between hard rubber and normal street pavement is about 0.85. On how steep a hill (maximum angle) can you leave a car parked?
A car travels to right at constant velocity. The net force on the car is
Friction provides the force needed for a car to travel around a flat, circular race track. Answer the following:
.49 N
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| Kinematics | Forces |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| 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] |
General Metric Conversion Chart
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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