| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | I = \frac{1}{3}ML^2 + M(L^2) | The moment of inertia I for the system consists of two parts: the inertia of the rod about the pivot, given by \frac{1}{3}ML^2 (since it’s pivoted at one end), and the inertia of the mass attached to the other end, calculated as M(L^2). |
| 2 | I = \frac{4}{3}ML^2 | Summing up the two contributions to the moment of inertia gives \frac{1}{3}ML^2 + ML^2 = \frac{4}{3}ML^2. |
| 3 | \tau = -MgL | The torque \tau generated by the mass at the end of the rod is calculated by the force due to gravity on the mass times the distance from the pivot. The negative sign indicates the torque acts to rotate the rod clockwise. |
| 4 | \alpha = \frac{\tau}{I} | The angular acceleration \alpha is found using Newton’s second law for rotation, which relates the torque on the system to its moment of inertia and angular acceleration. |
| 5 | \alpha = \frac{-MgL}{\frac{4}{3}ML^2} | Plugging in the values for \tau and I. |
| 6 | \alpha = \frac{-3g}{4L} | Upon simplifying, we find \alpha = \frac{-3g}{4L}. The negative sign shows the direction of the acceleration but for the magnitude we use \alpha = \frac{3g}{4L}. |
| 7 | (b) \: \frac{3g}{4L} | The correct option for the angular acceleration immediately after the rod is released is (b) \frac{3g}{4L}. |
In terms of evaluating the choices:
(a) \frac{g}{L} – Incorrect because it omits the contribution from the entire mass and length distribution.
(c) \frac{(m+1)g}{L} – Not suitable, incorrect dimensions and does not respect system specifications.
(d) \frac{3mg}{2L} – Incorrect as it miscalculates the distribution of mass.
(e) None of these – Not correct since one of the provided choices is indeed correct.
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A planet of constant mass orbits the sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet’s rotational kinetic energy about the sun’s center?
A uniform ladder of length L and weight W = 50 N rests against a smooth vertical wall. If the coefficient of static friction between the ladder and the ground is \mu = .4 .
A pulley has an initial angular speed of 12.5 rad/s and a constant angular acceleration of 3.41 rad/s2. Through what angle does the pulley turn in 5.26 s?
A rotating merry-go-round makes one complete revolution in 4.0 s. What is the linear speed and acceleration of a child seated 1.2 m from the center?
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during the entire 20-s interval? Assume constant angular acceleration while it is starting and stopping.
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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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