Step | Derivation/Formula | Reasoning |
---|---|---|
1 | v_{\text{critical}} = \sqrt{gR} | The critical speed v_{\text{critical}} is the minimum speed at the top of the loop which allows the roller coaster to remain in contact with the track and complete the loop without falling off. It is found by equating the gravitational force to the required centripetal force for circular motion at the top of the loop. |
2 | v = 2v_{\text{critical}} = 2\sqrt{gR} | It is given that the roller coaster crosses at twice the critical speed. Therefore, the actual speed v is twice the critical speed. |
3 | F_{\text{net}} = m \frac{v^2}{R} | The net force needed for circular motion at the top of the loop is given by the centripetal force formula where m is the mass of the roller coaster, v is the velocity, and R is the radius of the loop. |
4 | F_{\text{net}} = m \frac{(2\sqrt{gR})^2}{R} = 4mg | Substitute v = 2\sqrt{gR} into the centripetal force formula and simplify. |
5 | F_{\text{net}} = F_{\text{n}} + mg = 4mg | The net force F_{\text{net}} at the top of the loop is the sum of the normal force F_{\text{n}} exerted by the track on the roller coaster and the gravitational force mg . |
6 | F_{\text{n}} + mg = 4mg | Setting the net inward force (which includes the normal and gravitational forces) equal to the required centripetal force for motion at this speed. |
7 | F_{\text{n}} = 4mg \,-\, mg = 3mg | Solving for F_{\text{n}} by subtracting mg from each side. |
8 | \frac{F_{\text{n}}}{F_{\text{g}}} = \frac{3mg}{mg} = 3 | The ratio of the normal force to the gravitational force is \frac{F_{\text{n}}}{F_{\text{g}}} . Substituting the values from the previous step. |
9 | \frac{F_{\text{n}}}{F_{\text{g}}} = 3 | The final answer, indicating the ratio is 3, matching option (b). |
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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 curve with a radius of 125 m is properly banked for a car traveling 40 m/s. What must be the coefficient of static friction (µs) for a car not to skid on the same curve when traveling at 53 m/s?
Two identical object rests on a platform rotating at constant speed. Object A is at distance of half the platform’s radius from the center. Object B lays at edge of the platform. Assuming the platform continues rotating at the same speed, how does the centripetal force of the two objects compare?
Two wires are tied to the 500 g sphere shown below. The sphere revolves in a horizontal circle at a constant speed of 7.2 m/s. What is the tension in the upper wire? What is the tension in the lower wire?
A child has a toy tied to the end of a string and whirls the toy at constant speed in a horizontal circular path of radius R. The toy completes each revolution of its motion in a time period T. What is the magnitude of the acceleration of the toy (in terms of T, R, and g)?
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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_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 |
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