| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\theta_v = 90^\circ – 40^\circ = 50^\circ\] | The chain angle is given with the horizontal. For a conical pendulum, it is easiest to use the angle from the vertical, so convert using complementary angles: the angle from vertical is \(\theta_v\). |
| 2 | \[r = L\sin(\theta_v) = (8\,\text{m})\sin(50^\circ)\] | The person moves in a horizontal circle of radius \(r\). Geometry of the chain gives the horizontal radius as \(r\) equal to the horizontal component of \(L\). |
| 3 | \[\Sigma F_y = 0:\quad T\cos(\theta_v) – mg = 0\quad\Rightarrow\quad T\cos(\theta_v)=mg\] | Vertical acceleration is zero (the motion is horizontal). So the vertical component of tension balances weight. |
| 4 | \[\Sigma F_r = m\frac{v^2}{r}:\quad T\sin(\theta_v)=m\frac{v^2}{r}\] | The horizontal (radial) component of tension provides the centripetal force for circular motion. |
| 5 | \[\frac{T\sin(\theta_v)}{T\cos(\theta_v)}=\frac{m\frac{v^2}{r}}{mg}\quad\Rightarrow\quad \tan(\theta_v)=\frac{v^2}{rg}\] | Divide the radial equation by the vertical equation to eliminate \(T\) and \(m\), giving a direct relation between \(v\), \(r\), and \(\theta_v\). |
| 6 | \[v^2 = rg\tan(\theta_v)\] | Rearrange to solve for \(v\). This uses only algebra. |
| 7 | \[v = \sqrt{rg\tan(\theta_v)}=\sqrt{\big((8\,\text{m})\sin(50^\circ)\big)(9.8\,\text{m/s}^2)\tan(50^\circ)}\] | Substitute \(r=L\sin(\theta_v)\), \(g=9.8\,\text{m/s}^2\), and \(\theta_v=50^\circ\) into the speed expression. |
| 8 | \[v\approx \sqrt{\big(8\sin(50^\circ)\big)(9.8)\tan(50^\circ)}\approx \sqrt{71.6}\approx \boxed{8.46\,\text{m/s}}\] | Compute numerically: \(8\sin(50^\circ)\approx 6.13\,\text{m}\), \(\tan(50^\circ)\approx 1.19\), so \(v\approx 8.46\,\text{m/s}\). |
A Major Upgrade To Phy Is Coming Soon — Stay Tuned
We'll help clarify entire units in one hour or less — guaranteed.
A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.
A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed. What is the ratio of the normal force to the gravitational force?
An object moves at constant speed in a circular path of radius \( r \) at a rate of \( 1 \) revolution per second. What is its acceleration in terms of \(r\)?
A car is safely negotiating an unbanked circular turn at a speed of \(17 \, \text{m/s}\) on dry road. However, a long wet patch in the road appears and decreases the maximum static frictional force to one-fifth of its dry-road value. If the car is to continue safely around the curve, by what factor would the it need to change the original velocity?
Why do pilots sometimes black out while pulling out at the bottom of a dive?
Which of the following best explains why astronauts experience weightlessness while orbiting the earth?
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 \( 53 \) \( \text{m/s} \)?
Suppose you are a passenger traveling in car along a road that bends to the left. Why will you feel like you are being thrown against the door. What causes this force?
A satellite circling Earth completes each orbit in \(132 \, \text{minutes}\).
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 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 | |
|---|---|---|---|
| \( (a) \) | \( R\omega \) | \( R\omega^{2} \) | \( R\alpha \) |
| \( (b) \) | \( R\omega \) | \( R\alpha \) | \( R\omega^{2} \) |
| \( (c) \) | \( R\omega^{2} \) | \( R\alpha \) | \( R\omega \) |
| \( (d) \) | \( R\omega \) | \( R\omega^{2} \) | \( R\omega \) |
| \( (e) \) | \( R\omega^{2} \) | \( R\alpha \) | \( R\omega^{2} \) |
\(8.46\,\text{m/s}\)
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
