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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | $$ T_{1} – m_{1}g = m_{1}a $$ | This is Newton’s second law for mass \(m_{1}\) moving upward. |
| 2 | $$ m_{2}g – T_{2} = m_{2}a $$ | This is Newton’s second law for mass \(m_{2}\) moving downward. |
| 3 | $$ T_{1} = m_{1}g + m_{1}a \quad \text{and} \quad T_{2} = m_{2}g – m_{2}a $$ | Rearrange the equations to solve for the tensions in the string. |
| 4 | $$ (T_{2} – T_{1})R = I\left(\frac{a}{R}\right) $$ | This relates the net torque on the pulley to its moment of inertia \(I\) using the no‐slip condition \(\alpha = \frac{a}{R}\). |
| 5 | $$ T_{2} – T_{1} = \frac{I\,a}{R^{2}} $$ | Simplify the torque equation by dividing both sides by \(R\). |
| 6 | $$ (m_{2}g – m_{2}a) – (m_{1}g + m_{1}a) = $$$$(m_{2}-m_{1})g – (m_{2}+m_{1})a =$$$$\frac{I\,a}{R^{2}} $$ | Substitute the expressions for \(T_{1}\) and \(T_{2}\) into the torque equation. |
| 7 | $$ I = \frac{R^{2}}{a}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})a\Bigr] $$ | Rearrange the equation to solve for the moment of inertia \(I\). |
| 8 | $$ h = \frac{1}{2}at^{2} $$ | Use the kinematics relation for the heavy mass \(m_{2}\) falling a distance \(h\) from rest. |
| 9 | $$ a = \frac{2h}{t^{2}} $$ | Solve for the acceleration \(a\) from the kinematics equation. |
| 10 | $$ I = \frac{R^{2}}{\frac{2h}{t^{2}}}\Bigl[(m_{2}-m_{1})g – (m_{2}+m_{1})\frac{2h}{t^{2}}\Bigr] $$ | Substitute \(a = \frac{2h}{t^{2}}\) into the expression for \(I\). |
| 11 | $$ I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1}) $$ | Simplify the expression to obtain \(I\) solely in terms of \(m_{1}, m_{2}, R, h, t\) and \(g\). |
| 12 | $$ \boxed{I = \frac{R^{2}t^{2}}{2h}\Bigl[(m_{2}-m_{1})g\Bigr] – R^{2}(m_{2}+m_{1})} $$ | This is the final algebraic expression for the pulley’s moment of inertia. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | $$ \Delta x = R\theta $$ | This equation relates the linear displacement \(\Delta x\) to the angular displacement \(\theta\) of the pulley. |
| 2 | $$ h = R\theta $$ | Since the heavy mass \(m_{2}\) falls a distance \(h\), the length of the unwound rope is \(h\), which equals \(R\theta\). |
| 3 | $$ \theta = \frac{h}{R} $$ | Solve for the angular displacement \(\theta\) of the pulley. |
| 4 | $$ \boxed{\theta = \frac{h}{R}} $$ | This is the final expression for the total rotation of the pulley in radians. |
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Pulleys \( X \) and \( Y \) are each attached to a block by a string that wraps around the pulley. Both blocks are released and have the same linear acceleration \( a \). As the blocks fall, the pulleys rotate about their centers. Pulley \( Y \) has a larger radius than Pulley \( X \). How does the angular acceleration \( \alpha_X \) of Pulley \( X \) compare to the angular acceleration \( \alpha_Y \) of Pulley \( Y \)?
A miniature, solid globe with mass \( 0.25 \) \( \text{kg} \) and radius \( 0.10 \) \( \text{m} \) is spinning in place about a vertical axis with the equator horizontal, as shown. A point on the globe’s equator, represented by the dot in the figure, has a linear speed of \( 4.0 \) \( \text{m/s} \). The rotational inertia of a solid sphere of mass \( m \) and radius \( r \) is \( \tfrac{2}{5}mr^{2} \). The rotational kinetic energy of the globe is most nearly
A Christmas ornament made from a thin hollow glass sphere hangs from a thin wire of negligible mass. It is observed to oscillates with a frequency of \( 2.50 \) \( \text{Hz} \) in a city where \( g = 9.80 \) \( \text{m/s}^2 \). What is the radius of the ornament? The moment of inertia of the ornament is given by \( I = \frac{5}{3} mr^2 \).
A uniform stick has length \( L \). The moment of inertia about the center of the stick is \( I_0 \). A particle of mass \( M \) is attached to one end of the stick. The moment of inertia of the combined system about the center of the stick is
The angular velocity of a rotating disk of radius \(20 \, \text{cm}\) increases from \(1 \, \text{rad/s}\) to \(3 \, \text{rad/s}\) in \(0.5 \, \text{s}\). What is the linear tangential acceleration of a point on the rim of the disk during this time interval?
The diagram above shows a top view of a child of mass \(M\) on a circular platform of mass \(2M\) that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will increase the angular speed of the platform-child system and gives the correct reason why?
A boy and a girl are balanced on a massless seesaw. The boy has a mass of \(60 \, \text{kg}\) and the girl’s mass is \(50 \, \text{kg}\). If the boy sits \(1.5 \, \text{m}\) from the pivot point on one side of the seesaw, where must the girl sit on the other side for equilibrium?
When a fan is turned off, its angular speed decreases from \( 10 \) \( \text{rad/s} \) to \( 6.3 \) \( \text{rad/s} \) in \( 5.0 \) \( \text{s} \). What is the magnitude of the average angular acceleration of the fan?
A uniform copper disk of radius \( R \) has a moment of inertia \( I \) around an axis passing through the center of the disk perpendicular to its plane. If the radius of the disk were only \( \dfrac{R}{2} \), but the thickness were the same, what would be the moment of inertia in terms of \( I \)? Hint: The moment of inertia of a solid disk about its center is \(\frac{1}{2} M R^{2}\).
The system above is NOT balanced since \(m_2\) is twice the mass of \(m_1\). Which of the following changes would NOT balance the system so that there is 0 net torque? Assume the plank has no mass of its own.
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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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