| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) | \[\alpha = \frac{\Delta\omega}{\Delta t}=\frac{-30-60}{6-2}=\frac{-90}{4}=-22.5\,\text{rad\,/s}^2\] | The interval from \(t=2\,\text{s}\) to \(t=6\,\text{s}\) is a straight line, so the slope (constant) gives the instantaneous angular acceleration at any time in that portion, including \(t=4\,\text{s}.\) |
| (b) | \[v=r\,\omega = 0.25\,\text{m}\times 60\,\text{rad\,/s}=15\,\text{m\,/s}\] | At \(t=1\,\text{s}\) the graph is flat at \(\omega = 60\,\text{rad\,/s}.\) Linear (rim) speed is \(v=r\omega.\) |
| (c) | \[\Delta\theta_{0\!\to 2}=\omega\,\Delta t = 60\,\text{rad\,/s}\times 2\,\text{s}=120\,\text{rad}\] | From 0–2 s the angular velocity is constant, so the area under the \(\omega\)-vs-\(t\) graph (a rectangle) is \(\omega\Delta t.\) |
| (d) | \[\omega_f = \omega_i + \alpha\,\Delta t = 60 + (-22.5)(2)=15\,\text{rad\,/s}\] \[\Delta\theta_{2\!\to 4}=\tfrac12(\omega_i+\omega_f)\,\Delta t = \tfrac12(60+15)\times 2 = 75\,\text{rad}\] |
The segment 2–4 s lies on the linear portion with constant \(\alpha.\) Use kinematics (or trapezoid area) with \(\omega_i=60\,\text{rad\,/s}\) and \(\omega_f=15\,\text{rad\,/s}.\) |
| (e) | \[\begin{aligned} \Delta\theta_{0\!\to 2}&=120\\[4pt] \Delta\theta_{2\!\to 6}&=\tfrac12(60+(-30))(4)=60\\[4pt] \Delta\theta_{6\!\to 8}&=(-30)(2)=-60\\[4pt] \Delta\theta_{8\!\to 10}&=\tfrac12(-30+0)(2)=-30\\[4pt] \theta_{\text{total}}&=120+60-60-30=90\,\text{rad} \end{aligned}\] |
Sum the signed areas (trapezoids/rectangles) for each time interval. Positive areas correspond to counter-clockwise rotation; negative areas to clockwise. |
| (f) | \[\Delta x = r\,\theta_{\text{total}} = 0.25\,\text{m}\times 90\,\text{rad}=22.5\,\text{m}\] | For rolling without slipping, the center of mass translates a linear distance equal to \(r\,\Delta\theta.\) |
| (g) | \[a_{\text{tan}} = r\,|\alpha| = 0.25\,\text{m}\times 22.5\,\text{rad\,/s}^2 = 5.6\,\text{m\,/s}^2\] | Magnitude of tangential acceleration is the product of radius and magnitude of angular acceleration at \(t=4\,\text{s}.\) |
| (h) | \[a_{\text{tan}} = r\,\alpha = 0.25\times 0 = 0\,\text{m\,/s}^2\] \[v = r\omega = 0.25\times 60 = 15\,\text{m\,/s}\quad\Rightarrow\quad a_{c}=\frac{v^2}{r}=\frac{15^2}{0.25}=900\,\text{m\,/s}^2\] |
At \(t=1\,\text{s}\) the graph is flat, so \(\alpha=0\Rightarrow a_{\text{tan}}=0.\) Centripetal acceleration depends on instantaneous speed: \(a_c = v^2/r.\) |
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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 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.
A child of mass \( 3 \) \( \text{kg} \) rotates on a platform of \( 10 \) \( \text{kg} \). They start walking towards the center while the platform is rotating. Which of the following could possibly decrease the total angular momentum of the child-platform system?
The moment of inertia of a solid cylinder about its axis is given by \( 0.5MR^2 \). If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
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? \( I_{sphere} = \frac{2}{5}MR^2\), \( I_{cylinder} = \frac{1}{2}MR^2\), \( I_{pipe} = MR^2\)
A high-speed drill rotating counterclockwise at \( 2400 \) \( \text{rpm} \) comes to a halt in \( 2.5 \) \( \text{s} \).
What is the rotational inertia \( I \) of a disk with a radius \( R = 4 \) \( \text{m} \) and a mass \( 2 \) \( \text{kg} \)? The same disk is rotated around an axis that is \( 0.5 \) \( \text{m} \) from the center of the disk. What is the new rotational inertia \( I \) of the disk? What would the rotational inertia be if the disk axis was \( 3.75 \) \( \text{m} \) from the center?
A traffic light hangs from a pole as shown in the diagram. The uniform aluminum pole AB is of length \( 7.20 \) \( \text{m} \) and has a mass of \( 12.0 \) \( \text{kg} \). The mass of the traffic light is \( 21.5 \) \( \text{kg} \). The point C is located \( 3.80 \) \( \text{m} \) vertically above the pivot A. A massless horizontal cable CD is attached at C and connects to the pole at point D, where the pole makes an angle of \( 37^{\circ} \) with the cable.
Angular momentum cannot be conserved if
Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the second force is applied at \( 30^\circ \) to the plane of the door. Which force exerts the greater torque about the door hinge?
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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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