Step | Derivation/Formula | Reasoning |
---|---|---|
1 | x_{M_1} = \frac{L}{2} | The boy is sitting at one end of the seesaw, which places him at a distance of half the length of the plank (L) from the fulcrum. |
2 | x_{M_2} = \frac{L}{2} | 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 | x_M = \frac{L}{2} | The mass of the plank (M) is uniformly distributed, thus its center of mass is at the midpoint of the plank, which coincides with the fulcrum. |
4 | \tau_{M_1} = M_1 \cdot g \cdot \frac{L}{2} | Calculate the torque due to the boy’s mass at one end of the seesaw. Torque is given by \tau = r \times F where r is the distance from the pivot point and F is the force due to weight, which is M_1 \cdot g . |
5 | \tau_{M_2} = M_2 \cdot g \cdot \frac{L}{2} | Calculate the torque due to the girl’s mass at the other end. Similar to step 4, using the girl’s mass. |
6 | \tau_M = M \cdot g \cdot 0 | Calculate the torque due to the seesaw’s own mass. Since the seesaw’s center of mass is exactly at the fulcrum, the distance r is zero, thus the torque is zero. |
7 | \tau_{total} = \tau_{M_1} – \tau_{M_2} | 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 | M_1 \cdot g \cdot \frac{L}{2} = M_2 \cdot g \cdot \frac{L}{2} | For the seesaw to be balanced, the total torque must be 0. Setting the torques equal gives this balance condition. |
9 | M_1 = M_2 | Solve for the relationship between M_1 and M_2 . Since all other factors are equal and cancel out, the masses must be equal for balance. |
10 | M_1 = M_2 | 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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The figure shows a person’s foot. In that figure, the Achilles tendon exerts a force of magnitude F = 720 N. What is the magnitude of the torque that this force produces about the ankle joint?
A student is asked to design an experiment to determine the change in angular momentum of a disk that rotates about its center and the product of the average torque applied to the disk and the time interval in which the torque is exerted. A net force is applied tangentially to the surface of the disk. The rotational inertia of the disk about its center is I = MR^2. Which two of the following quantities should the student measure to determine the change in angular momentum of the disk after 10 s? Select two answers.
A solid sphere ( I = \frac{2}{5}MR^2) and a solid cylinder ( I = \frac{1}{2}MR^2), 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
A boy is sitting at a distance d_1 from the fulcrum, and girl is sitting at a distance d_2 from the fulcrum, with d_1 > d_2 . The seesaw is level, with the two ends at the same height. Derive an equation for the minimum mass of the seesaw that will keep it balanced with the two children on it.
An isolated spherical star of radius R_o , rotates about an axis that passes through its center with an angular velocity of \omega_o . Gravitational forces within the star cause the star’s radius to collapse and decrease to a value r_o <R_o , but the mass of the star remains constant. A graph of the star’s angular velocity as a function of time as it collapses is shown. Which of the following predictions is correct about the angular momentum L of the star immediately after the collapse?
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_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} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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