To solve the problem related to the balanced seesaw with a boy and a girl sitting on it, we adhere to the principles of torque and leverage. Here, the seesaw must balance so the torques due to the boy and girl must be equal in magnitude but opposite in direction.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \tau_{boy} = \tau_{girl} | This equation states the balancing condition where the torque (\tau) due to the boy must equal the torque due to the girl for the seesaw to be in equilibrium. |
| 2 | m_{boy} \cdot g \cdot d_1 = m_{girl} \cdot g \cdot d_2 | Torque (\tau) is calculated by the formula \tau = F \cdot d where F is the force (here, the weight of the children, m \cdot g) and d is the distance from the pivot. Here, g is the acceleration due to gravity, m_{boy} and m_{girl} are the masses of the boy and girl respectively, and d_1 and d_2 are their respective distances from the fulcrum. |
| 3 | \frac{m_{boy}}{m_{girl}} = \frac{d_2}{d_1} | Divide both sides of the equation by g \cdot d_1 \cdot d_2 to isolate the ratio of masses, which shows that the ratio of the boy’s mass to the girl’s mass is the inverse of their distances from the fulcrum. This ratio will ensure that their torques balance each other. |
| 4 | Mass of seesaw needed: m_{seesaw} \cdot g \cdot L = (m_{boy} + m_{girl}) \cdot g \cdot \frac{(d_2 – d_1)}{2} | We need to add the minimum mass of the seesaw to keep it balanced at the pivot point itself. Assuming the mass is evenly distributed, its leverage point would be at the center (\frac{L}{2} from the pivot). The seesaw’s mass should counteract any net torque resultant from the boy and girl’s differing distances from the pivot. Here, L is the total length of the seesaw. |
| 5 | m_{seesaw} = \frac{(m_{boy} + m_{girl}) \Big(\frac{(d_1 – d_2)}{2}\Big)}{L} | Re-arranging the equation to solve for m_{seesaw}. This formula calculates the minimum mass of the seesaw required to achieve balance. Note that since d_1 > d_2, (d_1 – d_2) will be positive, ensuring a positive mass for the seesaw. |
| 6 | m_{seesaw} = \frac{(m_{boy} + m_{girl}) \Big(\frac{(d_1 – d_2)}{2}\Big)}{L} | This is the final formula that yields the mass of the seesaw needed to balance with the boy and girl placed as described. |
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A 150-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope.
What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s?
Note: I_\text{disk} = \frac{1}{2}mr^2
A wheel of moment of inertia of 5.00 kg ·m2 starts from rest and accelerates under a constant torque of 3.00 N·m for 8.0 s. What is the wheel’s rotational kinetic energy at the end of 8.0 s?
A rod is initially at rest on a rough horizontal surface. Three forces are exerted on the rod with the magnitudes and directions shown in the figure. The force exerted in the center of the rod is an equidistant 0.5 m from both ends of the rod. If friction between the rod and the table prevents the rod from rotating, what is the magnitude of the torque exerted on the rod about its center from frictional forces?
Two masses, my = 32 kg and mg = 38 kg, are connected by a rope that hangs over a pulley. The pulley is a uniform cylinder of radius R = 0.311 m and mass 3.1 kg. Initially my is on the ground and mg rests 2.5 m above the ground. If the system is released, use conservation of energy to determine the speed of me just before it strikes the ground. Assume the pulley bearing is frictionless.
A disk of radius R = 0.5 cm rests on a flat, horizontal surface such that frictional forces are considered to be negligible. Three forces of unknown magnitude are exerted on the edge of the disk, as shown in the figure. Which of the following lists the essential measuring devices that, when used together, are needed to determine the change in angular momentum of the disk after a known time of 5.0 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_1m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu N |
| R = \frac{v_i^2 \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
| a_c = \frac{v^2}{r} | PE = mgh |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m v | \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 |
|---|
| F = -k x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
| M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
| M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kgm/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (Nm)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
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