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Step | Formula Derivation | Reasoning |
---|---|---|
1 | [katex] F_{\text{centripetal}} = \frac{mv^2}{r} [/katex] | Centripetal force formula, required for circular motion. |
2 | [katex] F_{\text{friction, dry}} = \mu_{\text{dry}} mg [/katex] | Static frictional force on dry road, where [katex] \mu_{\text{dry}} [/katex] is the coefficient of static friction on dry road and [katex] mg [/katex] is the gravitational force. |
3 | [katex] F_{\text{friction, wet}} = \frac{\mu_{\text{dry}}}{5} mg [/katex] | Static frictional force on wet road, reduced to one-fifth of its dry-road value. |
4 | [katex] F_{\text{centripetal, dry}} = F_{\text{friction, dry}} [/katex] | For safe negotiation of the turn on dry road, the centripetal force equals the static frictional force. |
5 | [katex] F_{\text{centripetal, wet}} = F_{\text{friction, wet}} [/katex] | For safe negotiation of the turn on wet road, the centripetal force equals the reduced static frictional force. |
6 | [katex] \frac{mv_{\text{dry}}^2}{r} = \mu_{\text{dry}} mg, \quad \frac{mv_{\text{wet}}^2}{r} = \frac{\mu_{\text{dry}}}{5} mg [/katex] | Applying the centripetal force formula for both dry and wet road conditions. |
7 | [katex] \frac{v_{\text{wet}}^2}{v_{\text{dry}}^2} = \frac{1}{5} [/katex] | Ratio of the squares of velocities on wet and dry roads. |
8 | [katex] \frac{v_{\text{wet}}}{v_{\text{dry}}} = \sqrt{\frac{1}{5}} [/katex] | Taking the square root to find the ratio of velocities. |
Let’s calculate the factor by which the velocity needs to change.
Step | Formula Derivation | Reasoning |
---|---|---|
8 | [katex] \frac{v_{\text{wet}}}{v_{\text{dry}}} \approx 0.447 [/katex] | Calculated ratio of velocities. |
To safely negotiate the turn on a wet road, the velocity of the car needs to be reduced by a factor of approximately [katex] \boxed{0.447}[/katex] or [katex] \boxed{\frac{4}{9}}[/katex] compared to its velocity on a dry road.
Just ask: "Help me solve this problem."
The International Space Station has a mass of 4.2 x105 kg and orbits Earth at a distance of 4.0 x102 km above the surface. Earth has a radius of 6.37 x106 m, and mass of 5.97 x1024 kg. Calculate the following:
What is a man’s apparent weight at the equator if his weight is 500 N? The earth’s radius is 6.37 x 106 m.
Friction provides the force needed for a car to travel around a flat, circular race track. Answer the following:
A person’s back is against the inner wall of spinning cylinder with no support under their feet. If the radius is R, find an expression for the minimum angular speed so the person does not slide down the wall. The coefficient of static friction is µs.
Note: If you haven’t studied angular velocity [katex] \omega [/katex] yet, just find the linear velocity v.
A [katex] 2.2 \times 10^{21} \, \text{kg}[/katex] moon orbits a distant planet in a circular orbit of radius [katex] 1.5 \times 10^8 \, \text{m}[/katex]. It experiences a [katex] 1.1 \times 10^{19} \, \text{N}[/katex] gravitational pull from the planet. What is the moon’s orbital period in earth days?
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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] |
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] | 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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