| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | \( F_{\text{centripetal}} = \frac{mv^2}{r} \) | Centripetal force formula, required for circular motion. |
| 2 | \( F_{\text{friction, dry}} = \mu_{\text{dry}} mg \) | Static frictional force on dry road, where \( \mu_{\text{dry}} \) is the coefficient of static friction on dry road and \( mg \) is the gravitational force. |
| 3 | \( F_{\text{friction, wet}} = \frac{\mu_{\text{dry}}}{5} mg \) | Static frictional force on wet road, reduced to one-fifth of its dry-road value. |
| 4 | \( F_{\text{centripetal, dry}} = F_{\text{friction, dry}} \) | For safe negotiation of the turn on dry road, the centripetal force equals the static frictional force. |
| 5 | \( F_{\text{centripetal, wet}} = F_{\text{friction, wet}} \) | For safe negotiation of the turn on wet road, the centripetal force equals the reduced static frictional force. |
| 6 | \( \frac{mv_{\text{dry}}^2}{r} = \mu_{\text{dry}} mg, \quad \frac{mv_{\text{wet}}^2}{r} = \frac{\mu_{\text{dry}}}{5} mg \) | Applying the centripetal force formula for both dry and wet road conditions. |
| 7 | \( \frac{v_{\text{wet}}^2}{v_{\text{dry}}^2} = \frac{1}{5} \) | Ratio of the squares of velocities on wet and dry roads. |
| 8 | \( \frac{v_{\text{wet}}}{v_{\text{dry}}} = \sqrt{\frac{1}{5}} \) | 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 | \( \frac{v_{\text{wet}}}{v_{\text{dry}}} \approx 0.447 \) | 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 \( \boxed{0.447}\) or \( \boxed{\frac{4}{9}}\) compared to its velocity on a dry road.
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A car with speed \( v \) and an identical car with speed \( 3v \) both travel the same circular section of an unbanked (flat) road. If the frictional force required to keep the faster car on the road without skidding is \( F \), then the frictional force required to keep the slower car on the road without skidding is
Riders in a carnival ride stand with their backs against the wall of a circular room of diameter \(8.0 \, \text{m}\). The room is spinning horizontally about an axis through its center at a rate of \(45 \, \text{rev/min}\) when the floor drops so that it no longer provides any support for the riders. What is the minimum coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall?
A group of astronauts in a spaceship are attempting to land on Mars. As they approach the planet, they begin to plan their descent to the surface.
A car is going through a dip in the road whose curvature approximates a circle of radius \( 200 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 20\% \) more than their normal weight \( (1.2\,W) \)?
An \(80 \, \text{kg}\) person sits on a swing ride that moves in a horizontal circle. The swing is suspended by a chain of length \(8 \, \text{m}\). While in motion, the chain makes a \(40^\circ\) angle with the horizontal. What is the speed of the person?
A pendulum consists of a ball of mass \( m \) suspended at the end of a massless cord of length \( L \). The pendulum is drawn aside through an angle of \( 60^\circ \) with the vertical and released. At the low point of its swing, the speed of the pendulum ball is
Two identical satellites are placed in orbit of two different planets. Satellite \(A\) orbits Mars, and Satellite \(B\) orbits Jupiter. The orbital speeds of each satellite are the same. Which satellite has a greater orbital radius?
A speed skater goes around a turn that has a radius of \(31 \, \text{m}\). The skater has a speed of \(14 \, \text{m/s}\) and experiences a centripetal force of \(460 \, \text{N}\). What is the mass of the skater?
Why do pilots sometimes black out while pulling out at the bottom of a dive?

A neighbor’s child wants to go to a carnival to experience the wild rides. The neighbor is worried about safety because one of the rides looks particularly dangerous. She knows that you have taken physics and so asks you for advice. The ride in question has a \(4 \, \text{kg}\) chair which hangs freely from a \(10 \, \text{m}\) long chain attached to a pivot on the top of a tall tower. When the child enters the ride, the chain is hanging straight down. The child is then attached to the chair with a seat belt and shoulder harness. When the ride starts up, the chain rotates about the tower. Soon the chain reaches its maximum speed and remains rotating at that speed, which corresponds to one rotation about the tower every \(3 \, \text{seconds}\). When you ask the operator, he says the ride is perfectly safe. He demonstrates this by sitting in the stationary chair. The chain creaks but holds, and he weighs \(90 \, \text{kg}\).
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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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