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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (a): Determine the final angular speed when power returns. | ||
| 1 | \( v_i = 500\,\text{rpm} \times \frac{2\pi\,\text{rad}}{60\,\text{s}} = \frac{500\times 2\pi}{60} = \frac{50\pi}{3}\,\text{rad/s} \) | Convert the initial speed from revolutions per minute to radians per second. |
| 2 | \( \Delta \theta = 200\,\text{rev} \times 2\pi = 400\pi\,\text{rad} \) | Calculate the total angular displacement during the 30 s power outage (each revolution is \(2\pi\) rad). |
| 3 | \( \Delta \theta = \frac{1}{2}(v_i + v_x)\,t \) | Use the kinematic equation for constant angular acceleration relating displacement, initial and final speeds over time \(t = 30\,s\). |
| 4 | \( v_x = \frac{2\Delta \theta}{t} – v_i = \frac{2(400\pi)}{30} – \frac{50\pi}{3} = \frac{800\pi}{30} – \frac{50\pi}{3} \) | Solve for the final angular speed \(v_x\) after 30 s. |
| 5 | \( \frac{800\pi}{30} = \frac{80\pi}{3}, \quad v_x = \frac{80\pi}{3} – \frac{50\pi}{3} = \frac{30\pi}{3} = 10\pi\,\text{rad/s} \) | Simplify the expression to obtain \(v_x\). This is equivalent to \(10\pi\,rad/s\) which can be converted to \(300\,rpm\) if desired. |
| 6 | \( \boxed{v_x = 10\pi\,\text{rad/s}} \) | Final answer for part (a): the flywheel spins at \(10\pi\,rad/s\) when power returns. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (b): Determine the time to stop and total revolutions if power did not return. | ||
| 1 | \( \alpha = \frac{v_x – v_i}{t} = \frac{10\pi – \frac{50\pi}{3}}{30} = \frac{\frac{30\pi – 50\pi}{3}}{30} = -\frac{20\pi}{90} = -\frac{2\pi}{9}\,\text{rad/s}^2 \) | Determine the constant angular deceleration \(\alpha\) using the change in angular velocity over 30 s. |
| 2 | \( 0 = v_i + \alpha T_{\text{stop}} \quad \Rightarrow \quad T_{\text{stop}} = -\frac{v_i}{\alpha} \) | Set the final angular velocity to zero to solve for the total stopping time \(T_{\text{stop}}\) from the moment of power failure. |
| 3 | \( T_{\text{stop}} = -\frac{\frac{50\pi}{3}}{-\frac{2\pi}{9}} = \frac{50\pi}{3} \times \frac{9}{2\pi} = 75\,\text{s} \) | Simplify to find that the flywheel takes 75 s to come to a complete stop. |
| 4 | \( \Delta \theta_{\text{total}} = \frac{v_i + 0}{2}T_{\text{stop}} = \frac{\frac{50\pi}{3}}{2} \times 75 = \frac{50\pi\times75}{6} = 625\pi\,\text{rad} \) | Calculate the total angular displacement using the average angular speed during deceleration. |
| 5 | \( \text{Revolutions} = \frac{625\pi}{2\pi} = 312.5\,\text{rev} \) | Convert radians to revolutions since \(2\pi\) rad correspond to one complete revolution. |
| 6 | \( \boxed{T_{\text{stop}} = 75\,\text{s} \quad \text{and} \quad \text{Total Revolutions} = 312.5\,\text{rev}} \) | Final answers for part (b): the flywheel stops in 75 s making a total of 312.5 revolutions. |
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The figure above shows a uniform beam of length \( L \) and mass \( M \) that hangs horizontally and is attached to a vertical wall. A block of mass \( M \) is suspended from the far end of the beam by a cable. A support cable runs from the wall to the outer edge of the beam. Both cables are of negligible mass. The wall exerts a force \( F_w \) on the left end of the beam. For which of the following actions is the magnitude of the vertical component of \( F_w \) smallest?
A construction worker spins a square sheet of metal of mass 0.040 kg with an angular acceleration of 10.0 rad/s2 on a vertical spindle (pin). What are the dimensions of the sheet if the net torque on the sheet is 1.00 N·m? Assume that the moment of inertia of a rectangle is [katex] I = \frac{1}{12}M(a^2+b^2) [/katex]
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]?
The angular velocity of a rotating disk of radius 20 cm increases from 1 rad/s to 3 rad/s in 0.5 s. What is the linear tangential acceleration of a point on the rim of the disk during this time interval?
A car is moving up the side of a circular roller coaster loop of radius \( 12 \) \( \text{m} \). The angular velocity is \( 1.8 \) \( \text{rad/s} \) and angular acceleration is \( -0.82 \) \( \text{rad/s}^2 \). The car is at the same elevation as the center of the loop. Find the magnitude and direction (relative to the horizontal) of the acceleration.
Part (a): \(\boxed{10\pi\,\text{rad/s}}\) (which is equivalent to 300 rpm).\nPart (b): \(\boxed{75\,\text{s}}\) to come to a complete stop with a total of \(\boxed{312.5\,\text{rev}}\).
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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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