| Step | Reasoning |
|---|---|
| Identify the total rotational inertia as the sum of the components about the same pivot point. \[ I_{total} = I_{rod} + I_{sensor} \] |
The system consists of two distinct parts (a uniform rod and a point mass), and rotational inertia is additive for objects sharing the same axis of rotation. |
| Determine the rotational inertia of the uniform rod about the off-center pivot using the parallel axis theorem. \[ I_{rod} = I_{cm} + Mh^2 \] |
The rod is uniform, so its center of mass is at its geometric center (\( L/2 \) from the end). The parallel axis theorem must be used to shift the inertia from the center of mass to the pivot point. |
| Identify the shift distance \( h \) for the rod and substitute into the rod’s inertia expression. \[ h = \left| \dfrac{L}{2} – d \right| \implies I_{rod} = \dfrac{1}{12}ML^2 + M\left(\dfrac{L}{2} – d\right)^2 \] |
The shift distance \( h \) is the physical distance between the rod’s center of mass and the pivot. Since the CM is at \( L/2 \) and the pivot is at \( d \) from the same end, the distance between them is the difference. |
| Calculate the rotational inertia of the point mass sensor about the pivot and combine the results. \[ I_{sensor} = md^2 \implies I_{total} = \dfrac{1}{12}ML^2 + M\left(\dfrac{L}{2} – d\right)^2 + md^2 \] |
For a point mass, the rotational inertia is \( mr^2 \) where \( r \) is the distance from the axis. The sensor is at the end, which is distance \( d \) from the pivot. |
Why each choice is correct or incorrect:
(A) Uses the formula for a rod pivoted at the end (\( \frac{1}{3}ML^2 \)), which only applies if \( d = 0 \) or \( d = L \).
(B) This is the correct answer.
(C) Misidentifies the shift distance in the parallel axis theorem as the distance from the pivot to the end of the rod (\( d \)) instead of the distance to the center of mass (\( L/2 – d \)).
(D) Uses the distance from the pivot to the wrong end of the rod (\( L – d \)) for the point mass calculation.
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Students are investigating the periodic motion of a physical pendulum to determine the acceleration due to gravity \(g\). They are provided with a uniform rigid rod of length \(L = 1.20 \text{ m}\) and mass \(M\). The rod has several small holes drilled along its length, allowing it to be pivoted about a frictionless axle. The distance from the center of mass of the rod to the pivot is \(x\).
When displaced by a small angle \(\theta\) and released, the rod oscillates. The theoretical period of oscillation \(T\) is given by the equation:
\[ T = 2\pi \sqrt{ \dfrac{\dfrac{1}{12}L^2 + x^2}{gx} } \]

A group of students is investigating the properties of a horizontal spring system. A cart of known mass \(M_c\) is attached to an ideal horizontal spring of unknown spring constant \(k\). The other end of the spring is attached to a fixed rigid wall. The cart rolls on a horizontal track with negligible friction. A motion sensor is positioned at the far end of the track to measure the speed of the cart. The students are provided with a set of identical metal blocks, each of known mass \(m_0\), which can be securely attached to the top of the cart. The students can pull the cart to stretch the spring by a distance \(D\) from its equilibrium position before releasing it from rest.

A solid cylinder of mass \(M\) and radius \(R\) is released from rest at the top of an incline of length \(L\) that makes an angle \(\theta\) with the horizontal, as shown in Figure 1. The cylinder rolls down the incline without slipping. As the cylinder moves, the air exerts a constant drag force of magnitude \(F_D\) on the cylinder, directed opposite to its velocity. The drag force acts entirely at the center of mass of the cylinder. The rotational inertia of a solid cylinder is \(I = \dfrac{1}{2}MR^2\).

A uniform rigid rod of mass \( M \) and length \( L \) hangs at rest from a fixed, frictionless pivot at its top end. A small block of mass \( m \) is sliding horizontally with speed \( v_0 \) toward the rod. The block strikes the rod at a distance \( d \) below the pivot and sticks to it. The rotational inertia of the rod about the pivot is \( \dfrac{1}{3}ML^2 \).

A uniform horizontal shelf of mass M is attached to a vertical wall by a hinge at its left end and is supported by a vertical wire at its right end. A student intends to use the rotational equilibrium condition \(\sum \tau = 0\) to determine the magnitude of the tension in the wire. The student knows the mass of the shelf but does not know the magnitude or direction of the force exerted by the hinge on the shelf. Which of the following describes the most useful location for the axis of rotation and provides a correct justification?

Block \(A\) of mass \(m_A\) rests on a rough horizontal table. Block \(B\) of mass \(m_B\) rests on top of Block \(A\). A light, inextensible string is attached to Block \(A\), passes over an ideal pulley at the edge of the table, and is attached to Block \(C\) of mass \(m_C\), which hangs vertically, as shown in Figure 1.
The coefficient of kinetic friction between Block \(A\) and the table is \(\mu_k\). The coefficient of static friction between Block \(A\) and Block \(B\) is \(\mu_s\). The system is released from rest, and Block \(B\) moves along with Block \(A\) without slipping. Assume air resistance is negligible.

Block A of mass \(m_A\) is traveling at speed \(v_0\) to the right on a horizontal, frictionless surface. An ideal, massless spring of spring constant \(k\) is attached to the right face of Block A. Block B of mass \(m_B\) is initially at rest on the surface. Block A collides with Block B, compressing the spring.

A solid sphere is launched across a rough horizontal surface with an initial translational velocity \(v_0\) to the right and zero initial angular velocity. As the sphere moves, the frictional force from the surface causes the sphere’s translational speed to decrease and its angular speed to increase until it eventually rolls without slipping. Which of the following correctly identifies the type and direction of the frictional force exerted by the surface on the sphere while it is sliding before it begins to roll without slipping?

A picture frame of mass Equivalence mass is suspended from a small, frictionless peg by a light, non-stretchable wire of total length \(L\). The ends of the wire are attached to two points at the top corners of the frame that are a distance \(w\) apart. The peg is positioned such that the two wire segments are of equal length and the system is in static equilibrium.
Which of the following expressions correctly represents the tension in the wire?

Two different spring-mass systems, A and B, oscillate in simple harmonic motion on a horizontal, frictionless surface. The potential energy \(U\) of each system as a function of displacement \(x\) from equilibrium is shown in the graph. The mass of the block in system B is twice the mass of the block in system A (\(m_B = 2m_A\)). What is the ratio \(T_B / T_A\) of the periods of the two systems?
B
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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] | 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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