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First, convert the rotational speed to radians per second: [katex]\omega = 45 , \frac{\text{rev}}{\text{min}} \times \frac{1 , \text{min}}{60 , \text{s}} \times \frac{2\pi , \text{radians}}{1 , \text{rev}}[/katex].
Note you can convert to linear velocity [katex] v [/katex] instead of using [katex] \omega [/katex], since [katex] v = \omega r [/katex], where [katex] r [/katex] is the radius of the circle.
The minimum coefficient of static friction is found using the balance of forces in the vertical direction, with the static frictional force providing the upward force to counteract gravity, and the centripetal force being provided by the normal force.
Step | Formula / Derivation | Reasoning |
---|---|---|
1 | [katex]a_c = \frac{v^2}{R}[/katex] | Centripetal acceleration formula. |
2 | [katex]v = R\omega[/katex] | Linear velocity in terms of radius and angular velocity. |
3 | [katex]a_c = R\omega^2[/katex] | Substitute [katex]v[/katex] into the centripetal acceleration formula. |
4 | [katex]F_c = m a_c = m R\omega^2[/katex] | Centripetal force provided by the normal force. |
5 | [katex]f_s = \mu_s N[/katex] | Static frictional force opposing gravity. |
6 | [katex]f_s = m g[/katex] | Static frictional force equals gravitational force to prevent sliding. |
7 | [katex]\mu_s = \frac{f_s}{N} = \frac{mg}{mR\omega^2}[/katex] | Substitute [katex]f_s[/katex] and [katex]N[/katex] with their equivalents. |
8 | [katex]\mu_s = \frac{g}{R\omega^2}[/katex] | Cancel [katex]m[/katex] from the equation. |
9 | [katex]\mu_s = \frac{g}{R\left(\frac{45}{60} \times 2\pi\right)^2}[/katex] | Substitute [katex]\omega[/katex] with its value in terms of rev/min to rad/s. |
10 | [katex]\mu_s = \frac{9.8}{4\left(\frac{3}{4} \pi\right)^2}[/katex] | Substitute given values for [katex]g[/katex] and [katex]R[/katex]. |
11 | [katex]\mu_s = \frac{9.8}{4\left(\frac{9}{16} \pi^2\right)}[/katex] | Simplify the expression for [katex]\omega^2[/katex]. |
12 | [katex]\boxed{\mu_s = \frac{9.8 \times 16}{4 \times 9 \pi^2}}[/katex] | Final expression for [katex]\mu_s[/katex]. |
Now let’s calculate the exact value for the minimum coefficient of static friction required.
The minimum coefficient of static friction required so that the rider does not slide down the wall is [katex] \boxed{\mu_s \approx 0.11} [/katex]
Just ask: "Help me solve this problem."
A \(5.0 \, \text{g}\) coin is placed \(15 \, \text{cm}\) from the center of a turntable. The coin has coefficients of static and kinetic friction of \(\mu_s = 0.80\) and \(\mu_k = 0.50\). The turntable slowly speeds up to \(60 \, \text{rpm}\). Does the coin slide off the turntable?
Which of the following do not affect the maximum speed that a car can drive in a circle? Choose both correct answers.
A car is going over the top of a hill whose curvature approximates a circle of radius \( 350 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 10\% \) less than their normal weight?
Why do you tend to slide across the car seat when the car makes a sharp turn?
Consider an object on a rotating disk at a distance \( r \) from its center, held in place on the disk by static friction. Which of the following statements is not true concerning this object?
[katex] \boxed{\mu_s \approx 0.11} [/katex]
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Kinematics | Forces |
---|---|
\(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
\(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
\(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
\(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
\(v^2 = v_f^2 \,-\, 2a \Delta x\) |
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] |
General Metric Conversion Chart
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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