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Unit 3 - Circular Motion

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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 [katex]v[/katex] on a curve of radius [katex]R[/katex] is [katex]F_c = \frac{mv^2}{R}[/katex]. The component of the normal force providing this centripetal force on a banked curve without friction is [katex]F_{c0} = \frac{mv_0^2}{R}[/katex].

The additional centripetal force required due to the increased speed must be provided by static friction [katex]f_s[/katex], which is [katex]\mu_s N[/katex], where [katex]N[/katex] is the normal force and [katex]\mu_s[/katex] is the coefficient of static friction.

Step Formula / Derivation Reasoning
1 [katex]F_{c0} = \frac{mv_0^2}{R}[/katex] Centripetal force for design speed.
2 [katex]F_{c} = \frac{mv^2}{R}[/katex] Centripetal force for actual speed.
3 [katex]f_s = F_{c} – F_{c0}[/katex] Additional force required from static friction.
4 [katex]f_s = \mu_s N[/katex] Static friction force.
5 [katex]N = mg \cos(\theta)[/katex] Normal force on banked curve, where [katex]\theta[/katex] is the banking angle.
6 [katex]\tan(\theta) = \frac{v_0^2}{Rg}[/katex] Relationship for a banked curve without friction.
7 [katex]\mu_s = \frac{f_s}{N}[/katex] Coefficient of static friction.
8 [katex]\mu_s = \frac{F_{c} – F_{c0}}{mg \cos(\theta)}[/katex] Combine steps 3, 4, and 5.
9 [katex]\mu_s = \frac{\frac{mv^2}{R} – \frac{mv_0^2}{R}}{mg \cos(\theta)}[/katex] Substitute [katex]F_{c}[/katex] and [katex]F_{c0}[/katex] from steps 1 and 2.
10 [katex]\mu_s = \frac{v^2 – v_0^2}{Rg \cos(\theta)}[/katex] Mass [katex]m[/katex] cancels out.
11 [katex]\mu_s = \frac{53^2 – 40^2}{125 \cdot 9.8 \cdot \cos(\theta)}[/katex] Substitute values for [katex]v[/katex], [katex]v_0[/katex], and [katex]R[/katex].
12 [katex]\mu_s = \frac{2809 – 1600}{1225 \cdot 9.8 \cdot \cos(\theta)}[/katex] Square the speeds.
13 [katex]\mu_s = \frac{1209}{1225 \cdot 9.8 \cdot \cos(\theta)}[/katex] Subtract the squares.
14 [katex]\mu_s = \frac{1209}{1225 \cdot 9.8 \cdot \frac{40^2}{125 \cdot 9.8}}[/katex] Substitute [katex]\cos(\theta)[/katex] from step 6.
15 [katex]\mu_s = \frac{1209 \cdot 125}{1225 \cdot 40^2}[/katex] Simplify.
16 [katex]\boxed{\mu_s = \frac{1209}{1960}}[/katex] Final calculation for [katex]\mu_s[/katex].

The coefficient of static friction required is approximately [katex]\frac{1209}{1960}[/katex], which can be simplified further if needed.

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[katex]\frac{1209}{1960} \approx .61 [/katex]

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Download Equation Sheet
KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque 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 MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[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
VariableSI 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]
VariableDerived 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]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. 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]

  3. 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]

  4. 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]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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