| Derivation/Formula | Reasoning |
|---|---|
| \[I = \frac{1}{2} M R^2\] | Moment of inertia for a uniform solid cylinder of mass \(M\) and radius \(R\). |
| \[\tau = T R\] | The tension \(T\) in the cord exerted at radius \(R\) produces torque \(\tau\). |
| \[\tau = I \alpha\] | Rotational form of Newton’s second law, relating torque to angular acceleration \(\alpha\). |
| \[\alpha = \frac{a}{R}\] | No slipping: linear acceleration \(a\) of the cord equals tangential acceleration \(R\alpha\). |
| \[T = \frac{I \alpha}{R}\] | Substitute \(\tau = T R\) into \(\tau = I\alpha\) and solve for tension. |
| \[T = \frac{1}{2} M a\] | Insert \(I = \tfrac{1}{2} M R^2\) and \(\alpha = a/R\); the \(R\) terms cancel. |
| \[m g – T = m a\] | Newton’s second law for the descending block; downward direction is positive. |
| \[m g – \tfrac{1}{2} M a = m a\] | Replace \(T\) with \(\tfrac{1}{2} M a\). |
| \[a = \frac{m g}{m + \tfrac{1}{2} M}\] | Solve the previous expression algebraically for \(a\). |
| \[a = \frac{(10)(9.8)}{10 + 50} = 1.633\, \text{m/s}^2\] | Insert \(m = 10\,\text{kg}\), \(M = 100\,\text{kg}\), and \(g = 9.8\,\text{m/s}^2\). |
| \[\boxed{a \approx 1.63\, \text{m/s}^2}\] | Final acceleration of the block. |
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One end of a string is wrapped around a pulley that is free to rotate with negligible friction about an axle at its center. The other end of the string is attached to a block. The block is released from rest and moves downward with constant acceleration. Which of the following correctly indicates whether the amount of work done on the pulley by the string during each successive complete rotation remains constant or increases, and provides a valid justification?
A pulley system consists of two blocks of mass \( 5 \) \( \text{kg} \) and \( 10 \) \( \text{kg} \), connected by a rope of negligible mass that passes over a pulley of radius \( 0.1 \) \( \text{m} \) and mass \( 2 \) \( \text{kg} \). The pulley is free to rotate about its axis. The system is released from rest, and the block of mass \( 10 \) \( \text{kg} \) starts to move downwards. Assume the pulley has a frictional force of \(5.7\) Newtons acting on the outer edge of the pulley.
A discus is held at the end of an arm that starts at rest. The average angular acceleration of \(54 \, \text{rad/s}^2 \) lasts for 0.25 s. The path is circular and has radius 1.1 m.
Note: A discuss is a heavy, flattened circular object for throwing.
When is the angular momentum of a system constant?
What is the ratio of the moment of inertia of a cylinder of mass \( m \) and radius \( r \) to the moment of inertia of a hoop of the same mass and same radius?
In an experiment, an external torque is applied to the edge of a disk of radius \( 0.5 \) \( \text{m} \) such that the edge of the disk speeds up as it continues to rotate. The tangential speed as a function of time is shown for the edge of the disk. The rotational inertia of the disk is \( 0.125 \) \( \text{kg} \cdot \text{m}^2 \). Can a student use the graph and the known information to calculate the net torque exerted on the edge of the disk?
A ball of radius \( r \) rolls on the inside of a circular track of radius \( R \). If the ball starts from rest at the left vertical edge of the track, what will be its speed when it reaches the lowest point of the track, rolling without slipping? For a solid spherical ball, the moment of inertia is \(\frac{2}{5} m r^2\).
During the experiment, students collect data about the angular momentum of a rigid, uniform spinning wheel about an axle as a function of time, which was used to create the graph that is shown. A frictional torque is exerted on the wheel. A student makes the following statement about the data. “The frictional torque exerted on the wheel is independent of the wheel’s angular speed.” Does the data from the graph support the student’s statement? Justify your selection.
An \( 80 \, \text{kg} \) block is placed \( 2 \, \text{m} \) away from the endpoint of a horizontal steel beam of length \( 6.6 \, \text{m} \) and mass \( 1,450 \, \text{kg} \). The plank makes contact with a vertical wall on one end (assume it does not slip). The other end of the beam is attached to a massless cable that makes an angle of \( 30^\circ \) with the horizontal and ties into the vertical wall as well. Calculate the (1) tension force in the cable and (2) the total force the wall exerts on the beam.
An object is moving in a horizontal circle at a constant speed. Which of the following correctly describes the linear and angular velocities of the object between any point along the circular path?
\(1.63\, \text{m/s}^2\)
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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] |
Metric Prefixes
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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