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| 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 uniform solid cylinder of mass [katex] M [/katex] and radius [katex] R [/katex] is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force [katex] F [/katex] , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle [katex] \theta [/katex], what is the kinetic energy of the cylinder in terms of [katex] F [/katex] and [katex] \theta [/katex]?
A car accelerates from \( 0 \) to \( 25 \) \( \text{m/s} \) in \( 5 \) \( \text{s} \). If the car’s tires have a diameter of \( 70 \) \( \text{cm} \), how many revolutions does a tire make while accelerating?
A hoop with a mass [katex]m[/katex] and unknown radius is rolling without slipping on a flat surface with an angular speed [katex]\omega[/katex]. The hoop encounters a hill and continues to roll without slipping until it reaches a maximum height [katex]h[/katex].
The object shown in the diagram below consists of a cylinder of mass \( 100 \) \( \text{kg} \) and radius \( 25.0 \) \( \text{cm} \) connected by four thin rods, each of mass \( 5.00 \) \( \text{kg} \) and length \( 0.75 \) \( \text{m} \), to a thin-outer ring of mass \( 20.0 \) \( \text{kg} \). A small chunk of metal of mass \( 1.00 \) \( \text{kg} \) is welded to the outer ring. Determine the moment of inertia of the entire assembly about the center of the inner cylinder, treating the metal chunk as a point mass. Hint: The moment of inertia of a disk about it center is \(\tfrac{1}{2} M R^2\), a thin rod about it center is \(\tfrac{1}{12}ML^2\), and a thin hoop about its center is \(I = MR^2\).
A solid ball and a cylinder roll down an inclined plane. Which reaches the bottom first?
When the speed of a rear-drive car is increasing on a horizontal road, what is the direction of the frictional force on the tires?
A disk of radius 35 cm rotates at a constant angular velocity of 10 rad/s. How fast does a point on the rim of the disk travel (in m/s)?
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
A grinding wheel is in the form of a uniform solid disk of radius \( 7.00 \) \( \text{cm} \) and mass \( 2.00 \) \( \text{kg} \). It starts from rest and accelerates uniformly under the action of the constant torque of \( 0.600 \) \( \text{N m} \) that the motor exerts on the wheel.
A man with mass \( m \) is standing on a rotating platform in a science museum. The platform can be approximated as a uniform disk of radius \( R \) that rotates without friction at a constant angular velocity \( \omega \). Two students are discussing what the man should do if he wishes to change the angular velocity of the platform.
Student A says that the man should run towards the center of the platform, because this will decrease the moment of inertia of the man-platform system. Since \( L \propto I \), the angular momentum will decrease proportionately and the platform will slow down.
Student B says that since the platform is rotating counterclockwise, the man should run in a clockwise direction to slow the platform down. His feet will exert a frictional torque on the platform, which will cause an angular acceleration of the man-platform system.
Explain what is correct and incorrect about each students statement if anything.
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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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