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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. |
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A solid sphere ([katex] I = \frac{2}{5}MR^2[/katex]) and a solid cylinder ([katex] I = \frac{1}{2}MR^2[/katex]), both uniform and of the same mass and radius, roll without slipping at the same forward speed. It is correct to say that the total kinetic energy of the solid sphere is
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?
A solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. Select all that would affect the angular velocity of the sphere at the bottom of the incline.
A system consists of two small disks, of masses m and 2m, attached to ends of a rod of negligible mass of length 3x. The rod is free to turn about a vertical axis through point P. The first mass, m, is located x away from point P, and therefore the other mass, of 2m, is 2x from point P. The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is . At time t = 0, the rod has an initial counterclockwise angular velocity ωi about P. The system is gradually brought to rest by friction.
Derive an expressions for the following quantities in terms of µ, m, x, g, and ωi.
A uniform solid cylinder of mass [katex] M [/katex] and radius [katex] R [/katex] is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force [katex] F [/katex] , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle [katex] \theta [/katex], what is the kinetic energy of the cylinder in terms of [katex] F [/katex] and [katex] \theta [/katex]?
Yes. Both children have identical masses. See working in explanation.
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Kinematics | Forces |
---|---|
[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
[katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
[katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
Circular Motion | Energy |
---|---|
[katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
[katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
[katex]KE_i + PE_i = KE_f + PE_f[/katex] |
Momentum | Torque and Rotations |
---|---|
[katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
[katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
[katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
Simple Harmonic Motion |
---|
[katex]F = -k x[/katex] |
[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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