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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 6.0-cm-diameter gear rotates with angular velocity \( \omega = \left(20-\frac {1}{2} t^2 \right) \, \text {rad/s} \), where \(t\) is in seconds. At \(t = 4.0 \, \text{s}\), what are
The angular velocity of an electric motor is \(\omega = \left(20 – \frac{1}{2} t^2 \right) \, \text{rad/s}\), where \(t\) is in seconds.
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 disk is initially rotating counterclockwise around a fixed axis with angular speed \( \omega_0 \). At time \( t = 0 \), the two forces shown in the figure above are exerted on the disk. If counterclockwise is positive, which of the following could show the angular velocity of the disk as a function of time?
The figure above shows a uniform beam of length \( L \) and mass \( M \) that hangs horizontally and is attached to a vertical wall. A block of mass \( M \) is suspended from the far end of the beam by a cable. A support cable runs from the wall to the outer edge of the beam. Both cables are of negligible mass. The wall exerts a force \( F_w \) on the left end of the beam. For which of the following actions is the magnitude of the vertical component of \( F_w \) smallest?
\(2.03 \text{ m/s}\)
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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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