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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]x_{M_1} = \frac{L}{2}[/katex] | The boy is sitting at one end of the seesaw, which places him at a distance of half the length of the plank ([katex]L[/katex]) from the fulcrum. |
| 2 | [katex]x_{M_2} = \frac{L}{2}[/katex] | The girl is sitting at the other end of the seesaw opposite to the boy, also at a distance of half the length of the plank from the fulcrum. |
| 3 | [katex]x_M = \frac{L}{2}[/katex] | The mass of the plank ([katex]M[/katex]) is uniformly distributed, thus its center of mass is at the midpoint of the plank, which coincides with the fulcrum. |
| 4 | [katex] \tau_{M_1} = M_1 \cdot g \cdot \frac{L}{2} [/katex] | Calculate the torque due to the boy’s mass at one end of the seesaw. Torque is given by [katex] \tau = r \times F [/katex] where [katex] r [/katex] is the distance from the pivot point and [katex] F [/katex] is the force due to weight, which is [katex] M_1 \cdot g [/katex]. |
| 5 | [katex] \tau_{M_2} = M_2 \cdot g \cdot \frac{L}{2} [/katex] | Calculate the torque due to the girl’s mass at the other end. Similar to step 4, using the girl’s mass. |
| 6 | [katex] \tau_M = M \cdot g \cdot 0 [/katex] | Calculate the torque due to the seesaw’s own mass. Since the seesaw’s center of mass is exactly at the fulcrum, the distance [katex] r [/katex] is zero, thus the torque is zero. |
| 7 | [katex] \tau_{total} = \tau_{M_1} – \tau_{M_2} [/katex] | Sum the torques. Torque due to the boy is assumed counterclockwise and positive, while that due to the girl is clockwise and negative (or vice versa depending on assignment). |
| 8 | [katex] M_1 \cdot g \cdot \frac{L}{2} = M_2 \cdot g \cdot \frac{L}{2} [/katex] | For the seesaw to be balanced, the total torque must be 0. Setting the torques equal gives this balance condition. |
| 9 | [katex] M_1 = M_2 [/katex] | Solve for the relationship between [katex] M_1 [/katex] and [katex] M_2 [/katex]. Since all other factors are equal and cancel out, the masses must be equal for balance. |
| 10 | [katex] M_1 = M_2 [/katex] | This shows that for the seesaw to remain balanced with a plank mass placed uniformly, the masses of the boy and girl must be equal. This is the condition for mechanical equilibrium. |
Just ask: "Help me solve this problem."
In the figure above, the marble rolls down the track and around a loop-the-loop of radius \( R \). The marble has mass \( m \) and radius \( r \). What minimum height \( h_{min} \) must the track have for the marble to make it around the loop-the-loop without falling off? Express your answer in terms of the variables \( R \) and \( r \).
A centrifuge rotor rotating at 9200 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 N·m. If the mass of the rotor is 3.10 kg and it can be approximated as a solid cylinder of radius 0.0710 m, through how many revolutions will the rotor turn before coming to rest? The moment of inertia of a cylinder is give by [katex] \frac{1}{2}mr^2 [/katex].
The angular velocity of an electric motor is \( \omega = \left(20 – \frac{1}{2} t^2 \right) \, \text{rad/s} \), where \(t\) is in seconds.
A disk increases from 2 complete revolutions in 2 seconds to 5 complete revolutions in 2 seconds. What is its average angular acceleration?
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]
Yes. Both children have identical masses. See working in explanation.
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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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