| 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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An object is experiencing a nonzero net force. Which of the following statements is most accurate?
An old record player could bring a disk up to its \(45\) RPM speed in less than a second. If the same size disk can also be brought up to a speed of \(75\) RPM in about the same amount of time on another player. Compare the torques exerted by each record player.
A boy is sitting at a distance \( d_1 \) from the fulcrum, and girl is sitting at a distance \( d_2 \) from the fulcrum, with \( d_1 > d_2 \). The seesaw is level, with the two ends at the same height. Derive an equation for the minimum mass of the seesaw that will keep it balanced with the two children on it.
| Wagon | Wheel Structure | Moment of Inertia | Wheel Mass | Wheel Radius |
|---|---|---|---|---|
| Wagon \(A\) | Solid disk | \[\frac{1}{2} M R^2\] | \[ 0.5 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
| Wagon \(B\) | Solid disk | \[\frac{1}{2} M R^2\] | \[ 0.2 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
| Wagon \(C\) | Hollow hoop | \[M R^2\] | \[ 0.1 \, \text{kg} \] | \[ 0.1 \, \text{m} \] |
Three wagons have identical total mass (including their wheels) and each has four wheels. However, the wheels on each wagon have different designs with varying mass distributions and radii as shown in a reference chart. When accelerating each wagon from a standstill to \( 10 \) \( \text{m/s} \), which wagon requires the most energy input?
A rod of length \( L \) is rotated about its center with \( I = \frac{ML^{2}}{12} \). What is the moment of inertia at a point \( \frac{L}{4} \) away from the center?
Four systems are in rotational motion. Which of the following combinations of rotational inertia and angular speed for each of the systems corresponds to the greatest rotational kinetic energy?
| System | Rotational Inertia | Angular Speed |
|---|---|---|
| A | \( I_0 \) | \( \omega_0 \) |
| B | \( I_0 \) | \( 4\, \omega_0 \) |
| C | \( 2 I_0 \) | \( 2\, \omega_0 \) |
| D | \( 6 I_0 \) | \( \omega_0 \) |
In both cases, a massless rod is supported by a fulcrum, and a \(200 \, \text{kg}\) hanging mass is suspended from the left end of the rod by a cable. A downward force \(F\) keeps the rod in rest. The rod in Case A is \(50 \, \text{cm}\) long, and the rod in Case B is \(40 \, \text{cm}\) long (each rod is marked at \(10 \, \text{cm}\) intervals). The magnitude of each vertical force \(F\) exerted on the rod will be
A \(350\ \text{g}\) ball is attached to the end of a thin, uniform rod of mass \(500\ \text{g}\) and length \(1.2\ \text{m}\). The system is rotated in a horizontal circle about the opposite end of the rod. Calculate the moment of inertia of the system about the axis of rotation. Hint: the moment of inertia of a thin rod about the end of the rod is \(I = \tfrac{1}{3} m L^2\).
A high-speed drill rotating counterclockwise at \( 2400 \) \( \text{rpm} \) comes to a halt in \( 2.5 \) \( \text{s} \).
An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his angular momentum about the axis of rotation?
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] |
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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