| Step | Formula and Calculations | Reasoning |
|---|---|---|
| 1 | \(v = \frac{2 \pi r}{T}\) | The velocity (v) is calculated using the formula for the circumference of a circle (2πr) divided by the period (T). |
| 2 | \(r = 2.6 , \text{km} = 2600 , \text{m}\) | Convert radius from kilometers to meters. |
| 3 | \(v = \frac{2 \pi \times 2600}{360}\) | Substitute the values of r (in meters) and T into the velocity formula. |
| 4 | \(a_c = \frac{v^2}{r}\) | Centripetal acceleration (a_c) is calculated using the square of the velocity divided by the radius. |
| 5 | \(a_c = \frac{(2 \pi \times 2600 / 360)^2}{2600}\) | Substitute the values of v and r into the centripetal acceleration formula. |
| 6 | \(a_c \approx 0.792 , \text{m/s}^2\) | Evaluate the centripetal acceleration to find its magnitude. |
The magnitude of the centripetal acceleration of the car is approximately 0.792 m/s².
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A comet of mass \( m_c = 3.2 \times 10^{14} \) \( \text{kg} \) is orbiting a star with mass \( m_s = 1.8 \times 10^{30} \) \( \text{kg} \). The comet’s orbit is elliptical. At its closest point, the comet is a distance \( r_1 = 8.3 \times 10^{10} \) \( \text{m} \) from the star, and at its farthest point, the comet is a distance \( r_2 = 4.9 \times 10^{11} \) \( \text{m} \) from the star. What is the change in the kinetic energy of the comet as it moves along its orbit from distance \( r_2 \) to distance \( r_1 \) from the star?
A curve with a radius of \( 125 \) \( \text{m} \) is properly banked for a car traveling \( 40 \) \( \text{m/s} \). What must be the coefficient of static friction \( (\mu_s) \) for a car not to skid on the same curve when traveling at \( 53 \) \( \text{m/s} \)?
A ball of mass \( M \) is attached to a string of length \( L \). It moves in a vertical circle and at the bottom the ball just clears the ground. The tension at the bottom of the path is \( 3 \) times the weight of the ball. Give all answers in terms of \( M \), \( L \), and \( g \).
A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?
A ball of mass \( m \) is fastened to a string. The ball swings at constant speed in a vertical circle of radius \( R \) with the other end of the string held fixed. Neglecting air resistance, what is the difference between the string’s tension at the bottom of the circle and at the top of the circle?
A 2.2 kg ball on the end of a 0.35 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 5.3 m/s. Find the tension in the string.
A planet of constant mass orbits the sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet’s rotational kinetic energy about the sun’s center?
The Moon does not crash into the Earth because:
A ball is attached to the end of a string. It is swung in a vertical circle of radius \( 0.33 \) \( \text{m} \). What is the minimum velocity that the ball must have to make it around the circle?
A car is safely negotiating an unbanked circular turn at a speed of 21 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a wet patch in the road decreases the maximum static frictional force to one-third its dry road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?
\(0.792 \, \text{m/s}^2\)
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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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