| 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. |
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A solid disk has a mass \( M \) and radius \( R \). What is the moment of inertia about an axis that is perpendicular to the plane of the disk and passes through its edge? Hint: the moment of inertia about the disk center is given as \(I_{center}=\frac{1}{2}M R^{2}\).
A motorcycle has tires with a diameter of \( 44.0 \) \( \text{cm} \). Cruising down the highway, they are rotating at \( 1150 \) \( \text{rpm} \) (revolutions per minute).
The graph above shows the angular velocity of a spinning wheel (radius = \( 25 \) \( \text{cm} \)) as a function of time.
Consider an object on a rotating disk at a distance \( r \) from its center, held in place on the disk by static friction. Which of the following statements is not true concerning this object?
The figure shows scale drawings of four objects, each of the same mass and uniform thickness, with the mass distributed uniformly. Which one has the greatest moment of inertia when rotated about an axis perpendicular to the plane of the drawing at point P?
A disk, a hoop, and a solid sphere are released at the same time at the top of an inclined plane. They are all uniform and roll without slipping. In what order do they reach the bottom?
\( \text{Solid sphere: } I = \frac{2}{5}mR^2, \quad \text{Solid disk: } I = \frac{1}{2}mR^2, \quad \text{Hoop: } I = mR^2 \)
Three masses are attached to a \( 1.5 \, \text{m} \) long massless bar. Mass 1 is \( 2 \, \text{kg} \) and is attached to the far left side of the bar. Mass 2 is \( 4 \, \text{kg} \) and is attached to the far right side of the bar. Mass 3 is \( 4 \, \text{kg} \) and is attached to the middle of the bar. At what distance from the far left side of the bar can a string be attached to hold the bar up horizontally?
A sphere starts from rest and rolls down an incline of height \( H = 1.0 \) \( \text{m} \) at an angle of \( 25^\circ \) with the horizontal, as shown above. The radius of the sphere \( R = 15 \) \( \text{cm} \), and its mass \( m = 1.0 \) \( \text{kg} \). The moment of inertia for a sphere is \( \frac{2}{5}mR^2 \). What is the speed of the sphere when it reaches the bottom of the plane?
A uniform meter stick has a mass of \( 45.0 \) \( \text{g} \) placed at the \( 20 \) \( \text{cm} \) mark. If a pivot is placed at the \( 42.5 \) \( \text{cm} \) mark and the meter stick remains horizontal in static equilibrium, what is the mass of the meter stick?
A string is wound tightly around a fixed pulley having a radius of 5.0 cm. As the string is pulled, the pulley rotates without any slipping of the string. What is the angular speed of the pulley when the string is moving at 5.0 m/s?
\(2.03 \text{ m/s}\)
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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] |
Metric Prefixes
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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