| Step | Reasoning |
|---|---|
| Identify the correct type of friction by evaluating the relative motion between the contact surfaces. | According to the definition of friction in Topic 2.7, kinetic friction occurs when two surfaces in contact are sliding relative to each other, while static friction occurs when there is no relative motion. |
| Determine the velocity of the contact point relative to the surface. \[ v_P = v_0 + 0 = v_0 \text{ (to the right)} \] |
The sphere is initially sliding with a center-of-mass velocity \(v_0\) and no rotation (\(\omega = 0\)). The velocity of the point of contact \(P\) is given by \(v_P = v_{cm} + v_{rot}\). Since \(v_{rot} = R\omega = 0\), the contact point moves at \(v_0\) to the right. |
| Apply the rule for the direction of kinetic friction. | Kinetic friction always acts in a direction that opposes the relative motion of the surfaces. Since the contact point of the sphere is moving to the right relative to the floor, the floor exerts a kinetic frictional force to the left. |
Why each choice is correct or incorrect:
(A) This is the correct answer; it accurately identifies that the relative motion requires kinetic friction and that the direction must oppose that motion.
(B) Incorrect because kinetic friction opposes the motion of the contact point (right) rather than supporting it; additionally, friction is not required to ‘keep an object moving’ according to Newton’s First Law.
(C) Incorrect because static friction only occurs when there is no relative motion between the contact point and the surface; here, the sphere is explicitly described as sliding.
(D) Incorrect because the eventual state of rolling without slipping does not define the current type of friction, which depends strictly on current relative motion; it also misidentifies the direction.
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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 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?

A space probe uses a thin uniform deployment boom of mass M and length L. A small scientific sensor of mass m is fixed to one end of the boom. The boom is attached to the probe at a pivot point located a distance d from the sensor, as shown in the figure, allowing the assembly to rotate in a horizontal plane. Which of the following is a correct expression for the total rotational inertia of the boom-sensor system about the pivot?
A
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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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