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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[10g – T_{2} = 10a\] | For the 10 kg block (which moves downward), the net force is its weight minus the rope tension \(T_{2}\) and equals \(10a\). |
| 2 | \[T_{1} – 5g = 5a\] | For the 5 kg block (which moves upward), the net force is the rope tension \(T_{1}\) minus its weight and equals \(5a\). |
| 3 | \[T_{2}=T_{1}\,e^{\mu\pi}\] | Since the rope slips on the pulley, the capstan relation applies over a contact angle of \(\pi\) radians with \(\mu=0.2\); thus, the tension on the heavy side is \(T_{2}=T_{1}\,e^{0.2\pi}\). |
| 4 | \[T_{1}=5g+5a\] | Solve the 5 kg block equation for \(T_{1}\). |
| 5 | \[10g -e^{0.2\pi}(5g+5a)=10a\] | Substitute \(T_{2}=e^{0.2\pi}(5g+5a)\) into the 10 kg block equation. |
| 6 | \[10g-5e^{0.2\pi}g = a(10+5e^{0.2\pi})\] | Distribute and gather like terms in \(a\) and the constants. |
| 7 | \[a = \frac{5g(2-e^{0.2\pi})}{5(2+e^{0.2\pi})} = \frac{g(2-e^{0.2\pi})}{2+e^{0.2\pi}}\] | Simplify the expression to obtain \(a\) in terms of \(g\) and \(e^{0.2\pi}\). |
| 8 | \[a \approx \frac{9.8(2-1.875)}{2+1.875} \approx \frac{9.8(0.125)}{3.875} \approx 0.32\,\text{m/s}^2\] | Using \(e^{0.2\pi}\approx1.875\) and \(g\approx9.8\,\text{m/s}^2\), we find the acceleration. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T_{1}=5g+5a\] | From the 5 kg block equation, solve for \(T_{1}\); substitute \(a\approx0.32\,\text{m/s}^2\) and \(g=9.8\,\text{m/s}^2\). |
| 2 | \[T_{1}\approx5(9.8)+5(0.32)=49+1.6\approx50.6\,\text{N}\] | Calculate \(T_{1}\) numerically. |
| 3 | \[T_{2}=T_{1}\,e^{0.2\pi}\approx50.6\times1.875\approx94.7\,\text{N}\] | Apply the capstan relation to find \(T_{2}\) on the 10 kg side. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[f = T_{2}-T_{1}\] | The difference in tensions is provided by the kinetic friction at the pulley–rope interface. |
| 2 | \[f \approx 94.7-50.6 \approx 44.1\,\text{N}\] | Substitute the computed tensions to find the frictional force magnitude. |
| 3 | \[\text{Direction: Counterclockwise}\] | Since the 10 kg block moves downward (causing clockwise tendency), the kinetic friction acts opposite to the rope’s slip, i.e. it exerts a counterclockwise force on the pulley. |
Just ask: "Help me solve this problem."
Wheels \( A \) and \( B \) are connected by a moving belt and are both free to rotate about their centers. The belt does not slip on the wheels. The radius of Wheel \( B \) is twice the radius of Wheel \( A \). Wheel \( A \) has constant angular speed \( \omega_A \) and Wheel \( B \) has constant angular speed \( \omega_B \). Which of the following correctly relates \( \omega_A \) and \( \omega_B \)?
Consider a uniform hoop of radius R and mass M rolling without slipping. Which is larger, its translational kinetic energy or its rotational kinetic energy?
A spinning ice skater on extremely smooth ice is able to control the rate at which she rotates by pulling in her arms. Which of the following statements are true about the skater during this process?
A small sphere hangs from a string attached to the ceiling of a uniformly accelerating train car. It is observed that the string makes an angle of \(37^\circ\) with respect to the vertical. The magnitude of the acceleration \(a\) of the train car is most nearly:
A horizontal uniform rod of length L and mass M is pivoted at one end and is initially at rest. A small ball of mass M (same masses) is attached to the other end of the rod. The system is released from rest. What is the angular acceleration of the rod just immediately after the system is released?
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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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