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| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[ mgh = \frac{1}{2}mv_x^2 + \frac{1}{2}I\left(\frac{v_x}{R}\right)^2 \] | Apply energy conservation for rolling without slipping. The gravitational potential energy \(mgh\) converts to translational kinetic energy \(\frac{1}{2}mv_x^2\) and rotational kinetic energy \(\frac{1}{2}I\omega^2\) with the no-slip condition \(\omega=\frac{v_x}{R}\). |
| 2 | \[ mgh = \frac{1}{2}mv_x^2\left(1+\frac{I}{mR^2}\right) \] | Substitute \(\omega=\frac{v_x}{R}\) and factor out \(\frac{1}{2}mv_x^2\) to combine the terms into a single expression. |
| 3 | \[ v_x^2 = \frac{2gh}{1+\frac{I}{mR^2}} \] | Solve for the translational speed \(v_x\) in terms of the moment of inertia factor \(\frac{I}{mR^2}\). |
| 4 | \[ \text{For a solid sphere: } \frac{I}{mR^2}=\frac{2}{5} \quad\Rightarrow\quad v_{x,\text{sphere}}^2=\frac{2gh}{1+\frac{2}{5}}=\frac{10gh}{7} \] | Substitute the moment of inertia for a solid sphere \(I=\frac{2}{5}mR^2\), leading to a higher translational speed. |
| 5 | \[ \text{For a solid cylinder: } \frac{I}{mR^2}=\frac{1}{2} \quad\Rightarrow\quad v_{x,\text{cylinder}}^2=\frac{2gh}{1+\frac{1}{2}}=\frac{4gh}{3} \] | Substitute the moment of inertia for a solid cylinder \(I=\frac{1}{2}mR^2\). Its greater inertia compared to the sphere reduces its translational speed. |
| 6 | \[ \text{For a hollow pipe: } \frac{I}{mR^2}=1 \quad\Rightarrow\quad v_{x,\text{pipe}}^2=\frac{2gh}{1+1}=gh \] | Substitute the moment of inertia for a hollow pipe \(I=mR^2\). The high inertia value further reduces the translational speed. |
| 7 | \[ \frac{10gh}{7} > \frac{4gh}{3} > gh \] | Compare the expressions for \(v_x^2\): The solid sphere has the largest value, so it reaches the bottom first. The cylinder and pipe are slower because more energy goes into rotation. |
| 8 | \[ \boxed{(a)\;\text{Sphere}} \] | The sphere reaches the bottom first. The cylinder and pipe are slower due to their higher moments of inertia; hence, they devote a larger portion of the energy to rotation. |
Just ask: "Help me solve this problem."
A point on the edge of a disk rotates around the center of the disk with an initial angular velocity of 3 rad/s clockwise. The graph shows the point’s angular acceleration as a function of time. The positive direction is considered to be counterclockwise. All frictional forces are considered to be negligible.
Suppose just two external forces act on a stationary, rigid object and the two forces are equal in magnitude and opposite in direction. Under what condition does the object start to rotate?
A rotating, rigid body makes 10 complete revolutions in 10 seconds. What is its average angular velocity?
A 5-meter long ladder is leaning against a wall, with the bottom of the ladder 3 meters from the wall. The ladder is uniform and has a mass of 20 kg. A person of mass 80 kg is standing on the ladder at a distance of 4 meters from the bottom of the ladder. The ladder makes an angle of 60 degrees with the ground. What is the force exerted by the wall on the ladder?
The elliptical orbit of a comet is shown above. Positions 1 and 2 are, respectively, the farthest and nearest positions to the Sun, and at position 1 the distance from the comet to the Sun is 10 times that at position 2. What is the ratio \(v_1\)/\(v_2\) of the speed of the comet at position 1 to the speed at position 2?
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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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