Step | Formula / Derivation | Reasoning |
---|---|---|
1 | F_c = \frac{mv^2}{R} | Centripetal force formula, where m is mass, v is speed, and R is radius. |
2 | v = \frac{2\pi R}{T} | Speed v is the circumference 2\pi R divided by period T . |
3 | T = \frac{1}{f} | Period T is the reciprocal of frequency f . |
4 | v = 2\pi R f | Substituting T = \frac{1}{f} into v = \frac{2\pi R}{T} . |
5 | F_c constant, R decreases | If R decreases and F_c is constant, v^2 must decrease to maintain the equation. |
6 | v decreases | Speed v must decrease because v^2 has decreased. |
7 | f increases | Since v decreases but R decreases more (being in the denominator), the overall effect is an increase in f . |
The correct answer is:
c. v decreases and f increases
In step 2, the relationship is derived from the definition of linear speed in circular motion, where the distance traveled in one revolution is the circumference of the circle. Substituting the period T with its reciprocal, the frequency f , directly gives us the relationship used in step 4. This process shows how the frequency of revolutions is inversely proportional to the period and directly proportional to the linear speed and radius of the path.
Phy can also check your working. Just snap a picture!
A 2 kg ball is swung in a vertical circle. The length of the string the ball is attached to is 0.7 m. It takes 0.4 s for the ball to travel one revolution ( assume ball travels at constant speed).
A person’s back is against the inner wall of 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 µs.
Note: If you haven’t studied angular velocity \omega yet, just find the linear velocity v.
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 (ac) of the rock?
A simple pendulum consists of a bob of mass 1.8 kg attached to a string of length 2.3 m. The pendulum is held at an angle of 30° from the vertical by a light horizontal string attached to a wall, as shown above.
A new car is tested on a 230-m-diameter track. If the car speeds up at a steady 1.4 \, m/s^2, how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?
By continuing you (1) agree to our Terms of Sale and Terms of Use 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_1m_2}{r^2} |
a = \frac{\Delta v}{\Delta t} | f = \mu N |
R = \frac{v_i^2 \sin(2\theta)}{g} |
Circular Motion | Energy |
---|---|
F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
a_c = \frac{v^2}{r} | PE = mgh |
KE_i + PE_i = KE_f + PE_f |
Momentum | Torque and Rotations |
---|---|
p = m v | \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 |
---|
F = -k x |
T = 2\pi \sqrt{\frac{l}{g}} |
T = 2\pi \sqrt{\frac{m}{k}} |
Constant | Description |
---|---|
g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
k | Spring constant, in \text{N/m} |
M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
Variable | SI Unit |
---|---|
s (Displacement) | \text{meters (m)} |
v (Velocity) | \text{meters per second (m/s)} |
a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
t (Time) | \text{seconds (s)} |
m (Mass) | \text{kilograms (kg)} |
Variable | Derived SI Unit |
---|---|
F (Force) | \text{newtons (N)} |
E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
P (Power) | \text{watts (W)} |
p (Momentum) | \text{kilogram meters per second (kgm/s)} |
\omega (Angular Velocity) | \text{radians per second (rad/s)} |
\tau (Torque) | \text{newton meters (Nm)} |
I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | 10^{-12} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
The most advanced version of Phy. Currently 50% off, for early supporters.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
A quick explanation
UBQ credits are specifically used to grade your FRQs and GQs.
You can still view questions and see answers without credits.
Submitting an answer counts as 1 attempt.
Seeing answer or explanation counts as a failed attempt.
Lastly, check your average score, across every attempt, in the top left.
MCQs are 1 point each. GQs are 1 point. FRQs will state points for each part.
Phy can give partial credit for GQs & FRQs.
Phy sees everything.
It customizes responses, explanations, and feedback based on what you struggle with. Try your best on every question!
Understand you mistakes quicker.
For GQs and FRQs, Phy provides brief feedback as to how you can improve your answer.
Aim to increase your understadning and average score with every attempt!
10 Free Credits To Get You Started
*Phy Pro members get unlimited credits
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.