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Part (a): Minimum coefficient
| Derivation/Formula | Reasoning |
|---|---|
| \[x_2 = \tfrac{L}{2}\cos\theta,\quad x_1 = d\cos\theta,\quad y_{\text{top}} = L\sin\theta\] | Set geometry: horizontal lever arms to the weights are \( \tfrac{L}{2}\cos\theta \) for the ladder’s center (mass \( m_2 \)) and \( d\cos\theta \) for the person (mass \( m_1 \)); the top contact is at height \( L\sin\theta \). |
| \[N_1(L\sin\theta) – m_2 g\left(\tfrac{L}{2}\cos\theta\right) – m_1 g(d\cos\theta) = 0\] | Torque about the bottom with \( \)counterclockwise positive: wall normal \( N_1 \) gives a positive moment \( N_1(L\sin\theta) \); weights \( m_2 g \) and \( m_1 g \) at offsets \( \tfrac{L}{2}\cos\theta \) and \( d\cos\theta \) give clockwise (negative) moments. |
| \[N_1 = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}\] | Solve the torque equation algebraically for \( N_1 \). |
| \[f – N_1 = 0\] | Horizontal force balance: the floor friction \( f \) (to the right) balances the wall’s leftward normal \( N_1 \), so \( f = N_1 \). |
| \[N_2 – (m_1 + m_2)g = 0\] | Vertical force balance: the ground normal \( N_2 \) supports the total weight \( (m_1+m_2)g \). |
| \[f = \mu_{\min} N_2\] | Impending slip condition at the threshold of motion defines \( \mu_{\min} \) via \( f = \mu_{\min} N_2 \). With \( f = N_1 \), one has \( \mu_{\min} = \tfrac{N_1}{N_2} \). |
| \[\mu_{\min} = \frac{\left(\tfrac{m_2 L}{2} + m_1 d\right)\cos\theta}{(m_1+m_2)L\sin\theta}\] | Substitute \( N_1 = \tfrac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta} \) and \( N_2 = (m_1+m_2)g \); the \( g \) cancels, yielding the simplified ratio. |
| \[\boxed{\mu_{\min} = \frac{\left(\tfrac{m_2 L}{2} + m_1 d\right)\cos\theta}{(m_1+m_2)L\sin\theta}}\] | Final expression for the minimum coefficient ensuring no slip. |
Part (b): Friction magnitude
| Derivation/Formula | Reasoning |
|---|---|
| \[\mu_s = \tfrac{3}{2}\,\mu_{\min}\] | Given that the available static friction coefficient \( \mu_s \) exceeds the minimum \( \mu_{\min} \) by a factor of \( \tfrac{3}{2} \). |
| \[f = N_1\] | In static equilibrium, the actual friction adjusts to balance horizontal forces; thus \( f \) equals the wall normal \( N_1 \), not \( \mu_s N_2 \) unless at the threshold. |
| \[N_1 = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}\] | Use the same torque result from part (a); it does not depend on \( \mu_s \). |
| \[\boxed{f = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}}\] | Substitute to obtain the friction magnitude; since \( \mu_s > \mu_{\min} \), this satisfies \( f \le \mu_s N_2 \) with margin. |
Just ask: "Help me solve this problem."
A net torque is applied to the edge of a spinning object as it rotates about its internal axis. The table shows the net torque exerted on the object at different instants in time. How can a student use the data table to determine the change in angular momentum of the object from \( 0 \) to \( 6 \) \( \text{s} \)? Justify your selection.
| Time \( (\text{s}) \) | Net Torque \( (\text{N} \cdot \text{m}) \) |
|---|---|
| 0 | 0 |
| 2 | 1.5 |
| 4 | 3.0 |
| 6 | 4.5 |
An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his angular momentum about the axis of rotation?
A person’s center of mass is easily found by having the person lie on a reaction board. A horizontal, \( 2.3 \) \( \text{m} \)-long, \( 6.1 \) \( \text{kg} \) reaction board is supported only at the ends, with one end resting on a scale and the other on a pivot. A \( 64 \) \( \text{kg} \) woman lies on the reaction board with her feet over the pivot. The scale reads \( 27 \) \( \text{kg} \). What is the distance from the woman’s feet to her center of mass? Express your answer with the appropriate units.
A high-speed flywheel in a motor is spinning at \( 500 \) \( \text{rpm} \) when a power failure suddenly occurs. The flywheel has a mass of \( 40 \) \( \text{kg} \) and a diameter of \( 75 \) \( \text{cm} \). The power is off for \( 30 \) \( \text{s} \) and during this time the flywheel slows due to friction in its axle bearings. During this time the flywheel makes \( 200 \) complete revolutions.
A friend is balancing a fork on one finger. Which of the following are correct explanations of how he accomplishes this? Select two answers.
\(\mu_{\min} = \frac{\left(\tfrac{m_2 L}{2} + m_1 d\right)\cos\theta}{(m_1+m_2)L\sin\theta}\)
\(f = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}\)
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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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