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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Real forces on the car: }\{\vec{W},\vec{N}\}\] | Because the road is slick and friction is negligible, there is no friction force. The only real forces are the car’s weight \(\vec{W}=M\vec{g}\) (downward) and the normal force \(\vec{N}\) (perpendicular to the road surface). |
| 2 | \[\vec{W}=Mg\,\text{(straight down)}\] | Weight always acts vertically downward with magnitude \(Mg\). |
| 3 | \[\vec{N}\perp\text{banked surface},\quad \angle(\vec{N},\text{vertical})=\theta\] | The normal force is perpendicular to the road. Since the road is banked at angle \(\theta\) above horizontal, the normal tilts by \(\theta\) from the vertical (toward the center of the circular path). |
| 4 | \[\text{Choose axes: }\hat{x}\text{ horizontal toward center (radial inward)},\;\hat{y}\text{ vertical upward}\] | This coordinate system matches the physics: centripetal acceleration is purely horizontal inward (radial), and there is no vertical acceleration because the car stays on the surface (constant height while going around). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\sum F_y = Ma_y = 0\] | There is no vertical acceleration for steady circular motion on the bank (the car does not accelerate upward or downward). |
| 2 | \[N\cos\theta – Mg = 0\] | Resolve \(\vec{N}\) into components using axes from part (a): vertical component is \(N\cos\theta\) upward, weight is \(Mg\) downward. |
| 3 | \[N\cos\theta = Mg\] | From \(\sum F_y=0\), the upward and downward forces balance. |
| 4 | \[\sum F_x = Ma_x = M\frac{v^2}{R}\] | Horizontal inward (radial) acceleration is centripetal with magnitude \(v^2/R\). |
| 5 | \[N\sin\theta = M\frac{v^2}{R}\] | The only horizontal inward force is the inward component of the normal force, \(N\sin\theta\). |
| 6 | \[\frac{N\sin\theta}{N\cos\theta} = \frac{M\frac{v^2}{R}}{Mg}\] | Divide the horizontal equation by the vertical equation to eliminate \(N\) and \(M\). |
| 7 | \[\tan\theta = \frac{v^2}{Rg}\] | Simplify: \(\sin\theta/\cos\theta=\tan\theta\) and \(\left(Mv^2/R\right)/(Mg)=v^2/(Rg)\). |
| 8 | \[v^2 = Rg\tan\theta\] | Solve algebraically for \(v^2\). |
| 9 | \[\boxed{v = \sqrt{Rg\tan\theta}}\] | This is the required speed for a frictionless banked curve; mass \(M\) cancels, so the speed does not depend on \(M\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[v = \sqrt{Rg\tan\theta}\] | Use the result from part (b). |
| 2 | \[v = \sqrt{(30.0\,\text{m})(9.80\,\text{m/s}^2)\tan(20.0^\circ)}\] | Substitute \(R=30.0\,\text{m}\), \(g=9.80\,\text{m/s}^2\), \(\theta=20.0^\circ\). (\(M\) is not needed because it cancels.) |
| 3 | \[v = \sqrt{(294\,\text{m}^2/\text{s}^2)\tan(20.0^\circ)}\] | Compute \((30.0)(9.80)=294\). |
| 4 | \[v \approx \sqrt{(294)(0.364)}\] | Use \(\tan(20.0^\circ)\approx 0.364\). |
| 5 | \[v \approx \sqrt{107}\] | Compute \((294)(0.364)\approx 107\). |
| 6 | \[\boxed{v \approx 10.3\,\text{m/s}}\] | Take the square root: \(\sqrt{107}\approx 10.3\), giving the car’s speed on the banked, frictionless curve. |
Just ask: "Help me solve this problem."
Consider a neutron star with a mass equal to the sun, a radius of 10 km, and a rotation period of 1.0 s. What is the radius of a geosynchronous orbit about the neutron star? The mass of the sun can be found in the formula sheet above.
A box is sliding down an incline at a constant speed of \( 2 \, \text{m/s} \). The angle of the incline is \( \theta \). The magnitude of the total of the opposing forces is \( 16 \, \text{N} \). Derive an equation for the force of gravity acting on the box.
A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.
A pulley system consists of two blocks of mass \( 5 \) \( \text{kg} \) and \( 10 \) \( \text{kg} \), connected by a rope of negligible mass that passes over a pulley of radius \( 0.1 \) \( \text{m} \) and mass \( 2 \) \( \text{kg} \). The pulley is free to rotate about its axis. The system is released from rest, and the block of mass \( 10 \) \( \text{kg} \) starts to move downwards. Assume the pulley has a frictional force of \(5.7\) Newtons acting on the outer edge of the pulley.
A block is given a brief push so that it slides up a ramp. After the block reaches its highest point, it slides back down, but the magnitude of its acceleration is less on the descent than on the ascent. Why?
\(v = \sqrt{Rg\tan\theta}\)
\(10.3\,\text{m/s}\)
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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] | |
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] |
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