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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."
Four systems are in rotational motion. Which of the following combinations of rotational inertia and angular speed for each of the systems corresponds to the greatest rotational kinetic energy?
| System | Rotational Inertia | Angular Speed |
|---|---|---|
| A | \( I_0 \) | \( \omega_0 \) |
| B | \( I_0 \) | \( 4\, \omega_0 \) |
| C | \( 2 I_0 \) | \( 2\, \omega_0 \) |
| D | \( 6 I_0 \) | \( \omega_0 \) |
Two uniform solid balls, one of radius \( R \) and mass \( M \), the other of radius \( 2R \) and mass \( 8M \), roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first?
One end of a string is wrapped around a pulley that is free to rotate with negligible friction about an axle at its center. The other end of the string is attached to a block. The block is released from rest and moves downward with constant acceleration. Which of the following correctly indicates whether the amount of work done on the pulley by the string during each successive complete rotation remains constant or increases, and provides a valid justification?
A ice skater that is spinning in circles has an initial rotational inertia Ii. You can approximate her shape to be a cylinder. She is spinning with velocity ωi. As she extends her arms she her rotational inertia changes by a factor of x and her angular velocity changes by a factor of y. Which one of the following options best describe x and y.
A rotating, rigid body makes 10 complete revolutions in 10 seconds. What is its average angular velocity?
\(2.03 \text{ m/s}\)
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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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