| Step | Formula / Calculation | Reasoning |
|---|---|---|
| 1 | [katex]F_{\text{centripetal}} = \frac{mv^2}{r}[/katex] | Centripetal force formula. |
| 2 | [katex]v = 14 , \text{m/s}, , r = 31 , \text{m}, , F_{\text{centripetal}} = 460 , \text{N}[/katex] | Given values for speed, radius, and centripetal force. |
| 3 | [katex]m = \frac{F_{\text{centripetal}} \times r}{v^2}[/katex] | Rearranging the centripetal force formula to solve for mass. |
| 4 | [katex]m = \frac{460 \times 31}{14^2}[/katex] | Substituting the given values into the equation. |
| 5 | [katex]m = \frac{460 \times 31}{196}[/katex] | Simplifying the denominator. |
| 6 | [katex]m = \frac{14260}{196}[/katex] | Calculating the numerator. |
| 7 | [katex]m \approx 72.76 , \text{kg}[/katex] | Final calculation of the mass. |
The mass of the skater is approximately 72.76 kg.
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A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tension in the cord connected to the ball be greatest?

Two wires are tied to the \(500 \, \text{g}\) sphere as shown above. The sphere revolves in a horizontal circle at a constant speed of \(7.2 \, \text{m/s}\). What is the tension in the upper wire? What is the tension in the lower wire?
The Moon does not crash into the Earth because:
The ultracentrifuge is an important tool for separating and analyzing proteins. Because of the enormous centripetal accelerations, the centrifuge must be carefully balanced, with each sample matched by a sample of identical mass on the opposite side. Any difference in the masses of opposing samples creates a net force on the shaft of the rotor, potentially leading to a catastrophic failure of the apparatus. Suppose a scientist makes a slight error in sample preparation and one sample has a mass \( 10 \) \( \text{mg} \) larger than the opposing sample.
If the samples are \( 12 \) \( \text{cm} \) from the axis of the rotor and the ultracentrifuge spins at \( 60000 \) \( \text{rpm} \), what is the magnitude of the net force on the rotor due to the unbalanced samples?
On a harsh winter day, a \( 1500 \) \( \text{kg} \) vehicle takes a circular banked exit ramp (radius \( R = 150 \) \( \text{m} \); banking angle of \( 10^\circ \)) at a speed of \( 30 \) \( \text{mph} \), since the speed limit is \( 35 \) \( \text{mph} \). However, the exit ramp is completely iced up (frictionless). To make matters worse, a wind is blowing parallel to the ramp in a downward direction. The wind exerts a force of \( 3000 \) \( \text{N} \). Under these conditions, can the driver continue to follow a safe horizontal circle on the exit ramp and stay below the speed limit?
To convert \( \text{mph} \) into \( \text{m/s} \), use \( 1 \) \( \text{mi} = 1607 \) \( \text{m} \) and \( 1 \) \( \text{hr} = 3600 \) \( \text{s} \).
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.
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?

An adult exerts a horizontal force on a swing that is suspended by a rope of length \( L \), holding it at an angle \( \theta \) with the vertical. The child in the swing has a weight \( W \) and dimensions that are negligible. In terms of \( W \) and \( \theta \), determine:
A person’s back is against the inner wall of a 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 \(\mu_s\).
If you haven’t studied angular velocity \(\omega\) yet, just find the minumum linear velocity \(v\).
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?
72.76 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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