| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \Delta L = \int_{0}^{6} \tau(t)\,dt \] | This is the fundamental relation in rotational dynamics stating that the change in angular momentum \(\Delta L\) is the time-integral (area under the curve) of the net torque \(\tau\) from \(t = 0\) to \(t = 6\) seconds. |
| 2 | \(\tau\ \text{vs.}\ t\ \text{graph}\) | The table provides net torque values at different times; plotting these yields a straight line indicating a linear increase in torque with time. |
| 3 | \(\tau(0) = 0,\; \tau(2) = 1.5,\; \tau(4) = 3.0,\; \tau(6) = 4.5\ \text{(in N\cdot m)}\) | The data shows that the net torque increases uniformly, so the graph from \(t=0\) to \(t=6\) seconds forms a right triangle. |
| 4 | \( \Delta L = \frac{1}{2}\,(\Delta t)\,(\tau_{\text{max}}) \) | Since the net torque increases linearly, the area under the \(\tau\) vs. \(t\) graph (which equals \(\Delta L\)) is the area of a triangle with base \(\Delta t = 6\) s and height \(\tau_{\text{max}} = 4.5\ \text{N\cdot m}\). |
| 5 | \( \Delta L = \frac{1}{2} \times 6\,\text{s} \times 4.5\,\text{N\cdot m} = 13.5\,\text{N\cdot m\cdot s} \) | This calculation gives the numerical value of the change in angular momentum over the 6-second interval. |
| 6 | Option (d): Graph the net torque vs. time and find the area under the curve. | This is the correct method because the area under the \(\tau\) vs. \(t\) graph directly represents the change in angular momentum \(\Delta L\), as derived above. |
| 7 | Incorrect Options: | – Option (a) incorrectly multiplies the maximum torque by the total time, ignoring that the torque is not constant. – Option (b) incorrectly divides the maximum torque by time. – Option (c) finds the slope, which does not yield the change in momentum because the slope represents the rate of change of torque, not the integrated effect over time. |
| 8 | \( \boxed{\Delta L = 13.5\,\text{N\cdot m\cdot s}} \) | This boxed result is the final change in angular momentum and confirms that option (d) is the proper procedure to use the data table. |
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If a constant net torque is applied to an object it will (select all that applies):
A uniform ladder of length \(L\) and weight \(W = 50 \, \text{N}\) rests against a smooth vertical wall. If the coefficient of static friction between the ladder and the ground is \(\mu = 0.4\).
A light string is attached to a massive pulley of known rotational inertia \( I_P \), as shown in the figure. A student must determine the relationship between the torque exerted on the pulley and the change in the pulley’s angular velocity when the torque is applied for \( 2.0 \) \( \text{s} \). In addition to a stopwatch to measure the time interval, what two measurements could the student make in order to determine the relationship? Select two answers.
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?
The figure shows a person’s foot. In that figure, the Achilles tendon exerts a force of magnitude F = 720 N. What is the magnitude of the torque that this force produces about the ankle joint?
A meter stick of mass 200 grams is balanced at the 40-cm mark when a 100-gram mass is suspended from the 10-cm mark. What is the distance from the pivot point to the center of mass of the meter stick? Give your answer in centimeters.
A string is wound tightly around a fixed pulley having a radius of 5.0 cm. As the string is pulled, the pulley rotates without any slipping of the string. What is the angular speed of the pulley when the string is moving at 5.0 m/s?
Which of the following must be true for an object at translational equilibrium?
In an experiment, an external torque is applied to the edge of a disk of radius \( 0.5 \) \( \text{m} \) such that the edge of the disk speeds up as it continues to rotate. The tangential speed as a function of time is shown for the edge of the disk. The rotational inertia of the disk is \( 0.125 \) \( \text{kg} \cdot \text{m}^2 \). Can a student use the graph and the known information to calculate the net torque exerted on the edge of the disk?
\( \boxed{\Delta L = 13.5\,\text{N\cdot m\cdot s}} \)
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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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