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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ U_i = m_z g (2.5) + m_y g (0) \] | Initial gravitational potential energy is only from \(m_z\) at a height of 2.5 m since \(m_y\) is on the ground. |
| 2 | \[ U_f = m_z g (0) + m_y g (2.5) \] | Final gravitational potential energy has \(m_y\) raised 2.5 m while \(m_z\) reaches the ground. |
| 3 | \[ \Delta U = U_f – U_i = (m_y – m_z)g (2.5) \] | The change in potential energy is the difference between the final and initial energies. Since \(m_y < m_z\) the result is negative, indicating energy conversion to kinetic energy. |
| 4 | \[ KE_{\text{total}} = \frac{1}{2}(m_z+m_y)v^2 + \frac{1}{2}I\omega^2 \] | The total kinetic energy is the sum of the translational kinetic energies of both masses and the rotational kinetic energy of the pulley. |
| 5 | \[ I = \frac{1}{2}MR^2 \quad \text{and} \quad \omega = \frac{v}{R} \] | For a uniform cylinder, the moment of inertia is \(\frac{1}{2}MR^2\), and the no-slip condition gives \(\omega = v/R\). |
| 6 | \[ KE_{\text{pulley}} = \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{4}Mv^2 \] | This expresses the pulley’s rotational kinetic energy in terms of \(v\). |
| 7 | \[ (m_z-m_y)g(2.5) = \frac{1}{2}(m_z+m_y)v^2 + \frac{1}{4}Mv^2 \] | Conservation of energy requires the loss in gravitational potential energy to equal the gain in kinetic energy. |
| 8 | \[ v^2 = \frac{(m_z-m_y)g(2.5)}{\frac{1}{2}(m_z+m_y) + \frac{1}{4}M} \] | Algebraically solving for \(v^2\) isolates the speed in terms of the given masses, gravity, and pulley mass. |
| 9 | \[ v = \sqrt{\frac{(38-32)(9.8)(2.5)}{\frac{1}{2}(32+38) + \frac{1}{4}(3.1)}} \] | Substitute \(m_z=38\) kg, \(m_y=32\) kg, and \(M=3.1\) kg. The numerator is \(6\cdot9.8\cdot2.5=147\) and the denominator is \(35+0.775=35.775\). |
| 10 | \[ v \approx \sqrt{\frac{147}{35.775}} \approx \sqrt{4.107} \approx 2.03 \text{ m/s} \] | Taking the square root gives the speed of \(m_z\) just before impact. |
| 11 | \[ \boxed{v = 2.03 \text{ m/s}} \] | This is the final answer using energy conservation. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ m_zg – T_z = m_za \] | For the falling mass \(m_z\), the net force is its weight minus the tension \(T_z\). |
| 2 | \[ T_y – m_yg = m_ya \] | For the rising mass \(m_y\), the net force is the tension \(T_y\) minus its weight. |
| 3 | \[ (T_z – T_y)R = I\alpha \] | The difference in tension produces a net torque on the pulley. With \(\alpha = \frac{a}{R}\) (no slip), this relates linear acceleration to angular acceleration. |
| 4 | \[ I = \frac{1}{2}MR^2 \quad \text{and} \quad \alpha = \frac{a}{R} \] | Substitute the moment of inertia for a uniform cylinder and express \(\alpha\) in terms of \(a\). |
| 5 | \[ T_z – T_y = \frac{1}{2}Ma \] | Simplify the torque equation using the expression for \(I\): \((T_z-T_y)R = \frac{1}{2}MR^2\cdot\frac{a}{R}\) leads to this equation. |
| 6 | \[ T_z = m_zg – m_za \quad \text{and} \quad T_y = m_yg + m_ya \] | Express tensions from the force equations for each mass. |
| 7 | \[ (m_z-g – m_z\,a) – (m_yg + m_ya) = (m_z-m_y)g – (m_z+m_y)a = \frac{1}{2}Ma \] | Subtracting the two tension expressions gives a relation between \(a\) and the masses. |
| 8 | \[ a = \frac{(m_z-m_y)g}{(m_z+m_y)+\frac{1}{2}M} \] | Rearrange the equation to solve for the linear acceleration \(a\). |
| 9 | \[ a = \frac{(38-32)\,9.8}{(38+32)+\frac{1}{2}(3.1)} = \frac{6\cdot9.8}{70+1.55} \] | Substitute \(m_z=38\) kg, \(m_y=32\) kg, and \(M=3.1\) kg. The denominator becomes \(70+1.55=71.55\) and the numerator is \(58.8\). |
| 10 | \[ a \approx \frac{58.8}{71.55} \approx 0.821 \text{ m/s}^2 \] | This yields the acceleration of the masses. |
| 11 | \[ v^2 = 2a\Delta x \quad \text{with} \quad \Delta x = 2.5 \text{ m} \] | Use the kinematic equation for constant acceleration, where \(\Delta x\) is the distance \(m_z\) falls. |
| 12 | \[ v = \sqrt{2(0.821)(2.5)} \approx \sqrt{4.105} \approx 2.03 \text{ m/s} \] | Taking the square root gives the final speed of \(m_z\), which verifies the answer from part (a). |
| 13 | \[ \boxed{v = 2.03 \text{ m/s}} \] | This confirms the speed obtained using the forces, torque, and kinematics method. |
Just ask: "Help me solve this problem."
A man with mass \( m \) is standing on a rotating platform in a science museum. The platform can be approximated as a uniform disk of radius \( R \) that rotates without friction at a constant angular velocity \( \omega \). Two students are discussing what the man should do if he wishes to change the angular velocity of the platform.
Student A says that the man should run towards the center of the platform, because this will decrease the moment of inertia of the man-platform system. Since \( L \propto I \), the angular momentum will decrease proportionately and the platform will slow down.
Student B says that since the platform is rotating counterclockwise, the man should run in a clockwise direction to slow the platform down. His feet will exert a frictional torque on the platform, which will cause an angular acceleration of the man-platform system.
Explain what is correct and incorrect about each students statement if anything.
A friend is balancing a fork on one finger. Which of the following are correct explanations of how he accomplishes this? Select two answers.
Which of the following must be true for an object at translational equilibrium?
A rod may freely rotate about an axis that is perpendicular to the rod and is along the plane of the page. The rod is divided into four sections of equal length of 0.2 m each, and four forces are exerted on the rod, as shown in the figure. Frictional forces are considered negligible. Which of the following describes an additional torque that must be applied in order to keep the rod from rotating?
A meter stick of mass 200 grams is balanced at the 40-cm mark when a 100-gram mass is suspended from the 10-cm mark. What is the distance from the pivot point to the center of mass of the meter stick? Give your answer in centimeters.
\(2.03 \text{ m/s}\)
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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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