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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.
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A car rounds a curve at a steady \( 50 \) \( \text{km/h} \). If it rounds the same curve at a steady \( 70 \) \( \text{km/h} \), will its acceleration be any different?
A car moves at constant speed in a circle of radius 75 m on a horizontal road. The coefficient of static friction is 0.62. Find the maximum speed the car can go without sliding.
A car travels at a constant speed around a circular track whose radius is \(2.6 \, \text{km}\). The car goes once around the track in \(360 \, \text{s}\). What is the magnitude of the centripetal acceleration of the car?
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 2.0 kg ball on the end of a 0.65 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 4.0 m/s. Find the tension in the string.
The occupants of a car traveling at a speed of \( 30 \) \( \text{m/s} \) note that on a particular part of a road their apparent weight is \( 15\% \) higher than their weight when driving on a flat road.
A \(2 \, \text{kg}\) ball is swung in a vertical circle. The length of the string the ball is attached to is \(0.7 \, \text{m}\). It takes \(0.4 \, \text{s}\) for the ball to travel one revolution (assume the ball travels at constant speed).
A point P is at a distance \( R \) from the axis of rotation of a rigid body whose angular velocity and angular acceleration are \( \omega \) and \( \alpha \) respectively. The linear speed, centripetal acceleration, and tangential acceleration of the point can be expressed as:
| Linear speed | Centripetal acceleration | Tangential acceleration | |
|---|---|---|---|
| \( (a) \) | \( R\omega \) | \( R\omega^{2} \) | \( R\alpha \) |
| \( (b) \) | \( R\omega \) | \( R\alpha \) | \( R\omega^{2} \) |
| \( (c) \) | \( R\omega^{2} \) | \( R\alpha \) | \( R\omega \) |
| \( (d) \) | \( R\omega \) | \( R\omega^{2} \) | \( R\omega \) |
| \( (e) \) | \( R\omega^{2} \) | \( R\alpha \) | \( R\omega^{2} \) |
A centripetal force of \( 5.0 \) newtons is applied to a rubber stopper moving at a constant speed in a horizontal circle. If the same force is applied, but the radius is made smaller, what happens to the speed, \( v \), and the frequency, \( f \), of the stopper?
Two satellites are in circular orbits around Earth. Satellite A has speed \(v_A\). Satellite B has an orbital radius nine times that of satellite A. What is the speed of satellite B?
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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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