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| 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. |
Just ask: "Help me solve this problem."
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 solid sphere \( I = 0.06 \, \text{kg} \cdot \text{m}^2 \) spins freely around an axis through its center at an angular speed of \( 20 \, \text{rad/s} \). It is desired to bring the sphere to rest by applying a friction force of magnitude \( 2.0 \, \text{N} \) to the sphere’s outer surface, a distance of \( 0.30 \, \text{m} \) from the sphere’s center. How much time will it take the sphere to come to rest?
A 5-meter long ladder is leaning against a wall, with the bottom of the ladder 3 meters from the wall. The ladder is uniform and has a mass of 20 kg. A person of mass 80 kg is standing on the ladder at a distance of 4 meters from the bottom of the ladder. The ladder makes an angle of 60 degrees with the ground. What is the force exerted by the wall on the ladder?
The diagram above shows a top view of a child of mass \(M\) on a circular platform of mass \(2M\) that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will increase the angular speed of the platform-child system and gives the correct reason why?
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] |
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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