| 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. |
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A platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the platform. A short time later, the disk and platform rotate together at the same angular velocity, as shown at right in the figure. How does the angular momentum of only the platform change, if at all, after the disk drops? And what is the best justification.
A discus is held at the end of an arm that starts at rest. The average angular acceleration of \(54 \, \text{rad/s}^2 \) lasts for 0.25 s. The path is circular and has radius 1.1 m.
Note: A discuss is a heavy, flattened circular object for throwing.
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?
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?
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?
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?
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 \(20 \, \text{kg}\). A person of mass \(80 \, \text{kg}\) is standing on the ladder at a distance of \(4\) meters from the bottom of the ladder. What is the force exerted by the wall on the ladder?
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?
The diagram above shows a top view of a child of mass \(M\) on a circular platform of mass \(5M\)that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will result in an increase in the total angular momentum of the child-platform system?
A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously at the top of an inclined plane and do not slip, which one will reach the bottom first? \( I_{sphere} = \frac{2}{5}MR^2\), \( I_{cylinder} = \frac{1}{2}MR^2\), \( I_{pipe} = MR^2\)
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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] |
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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