For the banked curve, the centripetal force necessary to make the turn is provided by the horizontal component of the normal force. When a car travels faster than the design speed, additional centripetal force must be provided by friction.
The centripetal force required for a car traveling at speed v on a curve of radius R is F_c = \frac{mv^2}{R}. The component of the normal force providing this centripetal force on a banked curve without friction is F_{c0} = \frac{mv_0^2}{R}.
The additional centripetal force required due to the increased speed must be provided by static friction f_s, which is \mu_s N, where N is the normal force and \mu_s is the coefficient of static friction.
Step | Formula / Derivation | Reasoning |
---|---|---|
1 | F_{c0} = \frac{mv_0^2}{R} | Centripetal force for design speed. |
2 | F_{c} = \frac{mv^2}{R} | Centripetal force for actual speed. |
3 | f_s = F_{c} – F_{c0} | Additional force required from static friction. |
4 | f_s = \mu_s N | Static friction force. |
5 | N = mg \cos(\theta) | Normal force on banked curve, where \theta is the banking angle. |
6 | \tan(\theta) = \frac{v_0^2}{Rg} | Relationship for a banked curve without friction. |
7 | \mu_s = \frac{f_s}{N} | Coefficient of static friction. |
8 | \mu_s = \frac{F_{c} – F_{c0}}{mg \cos(\theta)} | Combine steps 3, 4, and 5. |
9 | \mu_s = \frac{\frac{mv^2}{R} – \frac{mv_0^2}{R}}{mg \cos(\theta)} | Substitute F_{c} and F_{c0} from steps 1 and 2. |
10 | \mu_s = \frac{v^2 – v_0^2}{Rg \cos(\theta)} | Mass m cancels out. |
11 | \mu_s = \frac{53^2 – 40^2}{125 \cdot 9.8 \cdot \cos(\theta)} | Substitute values for v, v_0, and R. |
12 | \mu_s = \frac{2809 – 1600}{1225 \cdot 9.8 \cdot \cos(\theta)} | Square the speeds. |
13 | \mu_s = \frac{1209}{1225 \cdot 9.8 \cdot \cos(\theta)} | Subtract the squares. |
14 | \mu_s = \frac{1209}{1225 \cdot 9.8 \cdot \frac{40^2}{125 \cdot 9.8}} | Substitute \cos(\theta) from step 6. |
15 | \mu_s = \frac{1209 \cdot 125}{1225 \cdot 40^2} | Simplify. |
16 | \boxed{\mu_s = \frac{1209}{1960}} | Final calculation for \mu_s. |
The coefficient of static friction required is approximately \frac{1209}{1960}, which can be simplified further if needed.
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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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