| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[F_c=\frac{mv_x^2}{r}\] | A car moving in a circle needs centripetal force \(F_c\). The variables are mass \(m\), maximum speed \(v_x\), and radius \(r\). |
| 2 | \[F_{s,\max}=\mu_s mg\] | For a car driving without skidding, the needed centripetal force comes from static friction, not kinetic friction. The maximum static friction depends on the coefficient of static friction \(\mu_s\), mass \(m\), and gravitational acceleration \(g\). |
| 3 | \[\frac{mv_x^2}{r}=\mu_s mg\] | At the maximum speed before slipping, the required centripetal force equals the maximum static friction force. |
| 4 | \[\frac{v_x^2}{r}=\mu_s g\] | The mass \(m\) cancels from both sides, so the car’s mass does not affect the maximum speed. |
| 5 | \[v_x=\sqrt{\mu_s gr}\] | The maximum speed depends on \(\mu_s\), \(g\), and \(r\). It does not depend on the coefficient of kinetic friction \(\mu_k\) or the mass \(m\). |
| 6 | \[\boxed{\text{B, D}}\] | Choice \(\text{B}\), the coefficient of kinetic friction, does not affect the maximum speed because the tires are not sliding during normal circular motion. Choice \(\text{D}\), the mass of the car, does not affect the maximum speed because \(m\) cancels. Choices \(\text{A}\), \(\text{C}\), and \(\text{E}\) are incorrect because \(g\), \(\mu_s\), and \(r\) appear in \(v_x=\sqrt{\mu_s gr}\). |
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A concrete highway curve of radius \(60.0 \, \text{m}\) is banked at a \(12.0^\circ\) angle. What is the maximum speed with which a \(1300 \, \text{kg}\) rubber-tired car can take this curve without sliding? (Take the static coefficient of friction of rubber on concrete to be \(1.0\).)
Refer to the diagram above and solve all equations in terms of \(R\), \(M\), \(k\), and constants.
Which of the following best explains why astronauts experience weightlessness while orbiting the earth?
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 rock is whirled on the end of a string in a horizontal circle of radius \(R\) with a constant period \(T\). If the radius of the circle is reduced to \(R/3\), while the period remains \(T\), what happens to the centripetal acceleration (\(a_c\)) of the rock?
Consider an object on a rotating disk at a distance \( r \) from its center, held in place on the disk by static friction. Which of the following statements is not true concerning this object?
A satellite circling Earth completes each orbit in \(132 \, \text{minutes}\).
In the figure above, the marble rolls down the track and around a loop-the-loop of radius \( R \). The marble has mass \( m \) and radius \( r \). What minimum height \( h_{min} \) must the track have for the marble to make it around the loop-the-loop without falling off? Express your answer in terms of the variables \( R \) and \( r \).
A conical pendulum is formed by attaching a ball of mass \( m \) to a string of length \( \ell \), then allowing the ball to move in a horizontal circle of radius \( r \). The following figure shows that the string traces out the surface of a cone, hence the name.
A conical pendulum is formed by attaching a ball of mass \( m \) to a string of length \( L \), then allowing the ball to move in a horizontal circle of radius \( R \).
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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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