| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\tau_{10}=rF\sin\theta\] | The torque from the \(10\,\text{N}\) force equals the lever arm \(r=1\,\text{m}\) (midpoint) times the force times \(\sin\theta\) where \(\theta=37^{\circ}\) is the angle between the force and the rod. |
| 2 | \[\tau_{10}=1\times10\times\sin37^{\circ}=10(0.6)=6\,\text{N\,\cdot m}\] | Using \(\sin37^{\circ}\approx0.6\), the torque produced by the \(10\,\text{N}\) force is \(6\,\text{N\,m}\). |
| 3 | \[\tau_{F}=\tau_{10}\] | For rotational equilibrium, the net torque about the pivot must be zero, so the magnitude of the torque from force \(F\) must equal that from the \(10\,\text{N}\) force. |
| 4 | \[\boxed{\tau_{F}=6\,\text{N\,\cdot m}}\] | Thus the required torque produced by force \(F\) is \(6\,\text{N\,m}\). |
| 5 | \[3\,\text{N\,\cdot m}\] | Option (a) halves the correct torque; it ignores the correct trigonometric factor. |
| 6 | \[8\,\text{N\,\cdot m}\] | Option (c) uses \(\sin37^{\circ}\approx0.8\) instead of \(0.6\), overestimating the torque. |
| 7 | \[12\,\text{N\,\cdot m}\] | Option (d) doubles the actual torque, as if the force acted at the free end instead of the midpoint. |
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Three forces of equal magnitude are applied to a \(3 \, \text{m} \times 2 \, \text{m}\) rectangle. Force \(F_1\) and \(F_2\) act at \(45^\circ\) angles to the vertical as shown, while \(F_3\) acts horizontally.
In short:
\(F_1\): applied at \((0, -2)\), direction SW \(45^\circ\)
\(F_2\): applied at \((2, -2)\), direction NW \(45^\circ\)
\(F_3\): applied at \((3, -1)\), direction east
Points of rotation: \(A = (0, 0)\), \(B = (0, -1)\), \(C = (1, -1)\)
A motorcycle has tires with a diameter of \( 44.0 \) \( \text{cm} \). Cruising down the highway, they are rotating at \( 1150 \) \( \text{rpm} \) (revolutions per minute).
A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously at the top of an inclined plane and do not slip, which one will reach the bottom first? [katex] I_{sphere} = \frac{2}{5}MR^2[/katex], [katex] I_{cylinder} = \frac{1}{2}MR^2[/katex], [katex] I_{pipe} = MR^2[/katex]
During the experiment, students collect data about the angular momentum of a rigid, uniform spinning wheel about an axle as a function of time, which was used to create the graph that is shown. A frictional torque is exerted on the wheel. A student makes the following statement about the data. “The frictional torque exerted on the wheel is independent of the wheel’s angular speed.” Does the data from the graph support the student’s statement? Justify your selection.
A solid sphere is rotating about an axis through its center at a constant rotation rate. Another hollow sphere of the same mass and radius is rotating about its axis through the center at the same rotation rate. Which sphere has a greater rotational kinetic energy?
A meter stick of mass [katex] .2 [/katex] kg is pivoted at one end and supported horizontally. A force of [katex] 3 [/katex] N downwards is applied to the free end, perpendicular to the length of the meter stick. What is the net torque about the pivot point?
Two uniform disks have the same mass but different radii: disk \( 1 \) has a radius \( R \) and disk \( 2 \) has a radius \( 2R \). What is the ratio of the moment of inertia of the first disk to the second disk?
Old-fashioned clocks and watches have an hour hand, a minute hand and a second hand. What is the angular frequency of the second hand?
A massless rigid rod of length [katex]3d[/katex] is pivoted at a fixed point [katex]W[/katex], and two forces each of magnitude [katex]F[/katex] are applied vertically upward as shown above. A third vertical force of magnitude [katex]F[/katex] may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude [katex]F[/katex] is applied at point?
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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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