| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[a = \frac{g \sin\theta}{1 + \frac{I}{mR^2}}\] | For rolling without slipping, translational and rotational dynamics give this general acceleration down an incline. |
| 2 | \[I_{\text{ball}} = \frac{2}{5} m R^2\] | Moment of inertia of a solid sphere about its center. |
| 3 | \[a_{\text{ball}} = \frac{g \sin\theta}{1 + \frac{2}{5}} = \frac{5}{7} g \sin\theta\] | Substitute the sphere’s inertia into the general formula. |
| 4 | \[I_{\text{cyl}} = \frac{1}{2} m R^2\] | Moment of inertia of a solid cylinder about its center. |
| 5 | \[a_{\text{cyl}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{2}{3} g \sin\theta\] | Substitute the cylinder’s inertia into the general formula. |
| 6 | \[a_{\text{ball}} > a_{\text{cyl}}\] | Numerically, \(\frac{5}{7} \approx 0.714\) and \(\frac{2}{3} \approx 0.667\); the ball accelerates more. |
| 7 | \[\text{time} \propto \sqrt{\frac{2\,\Delta x}{a}}\] | For the same displacement \(\Delta x\), greater acceleration means less travel time. |
| 8 | \[\text{Ball reaches first}\] | Therefore the solid ball gets to the bottom before the cylinder—choice (a). |
| 9 | – | (b) is false because the cylinder’s lower acceleration makes it slower; (c) is false because their accelerations differ; (d) and (e) are false since mass and diameter cancel in the ratio \(I/(mR^2)\). |
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A system consists of two small disks, of masses \( m \) and \( 2m \), attached to a rod of negligible mass of length \( 3l \) as shown above. The rod is free to turn about a vertical axis through point \( P \). The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is \( \mu \). At time \( t = 0 \), the rod has an initial counterclockwise angular velocity \( \omega_0 \) about \( P \). The system is gradually brought to rest by friction. Develop expressions for the following quantities in terms of \( \mu \), \( m \), \( l \), \( g \), and \( \omega_0 \).
Suppose a solid uniform sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The angular velocity of the sphere at the bottom of the incline depends on
A rod is initially at rest on a rough horizontal surface. Three forces are exerted on the rod with the magnitudes and directions shown in the figure. The force exerted in the center of the rod is an equidistant 0.5 m from both ends of the rod. If friction between the rod and the table prevents the rod from rotating, what is the magnitude of the torque exerted on the rod about its center from frictional forces?
A \(5\)-meter long ladder is leaning against a wall, with the bottom of the ladder \(3\) meters from the wall. The ladder is uniform and has a mass of \(20 \, \text{kg}\). A person of mass \(80 \, \text{kg}\) is standing on the ladder at a distance of \(4\) meters from the bottom of the ladder. What is the force exerted by the wall on the ladder?
The driver of a car traveling at \( 30.0 \) \( \text{m/s} \) applies the brakes and undergoes a constant negative acceleration of \( 2.00 \) \( \text{m/s}^2 \). How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of \( 0.300 \) \( \text{m} \)?
A solid sphere of mass \( M \) and radius \( R \) rolls without slipping down an inclined plane starting from rest. Select all that would affect the angular velocity of the sphere at the bottom of the incline.
Two uniform disks have the same mass but different radii: disk \( 1 \) has a radius \( R \) and disk \( 2 \) has a radius \( 2R \). What is the ratio of the moment of inertia of the first disk to the second disk?
Two thin coins are made from identically the same metal, but one coin has triple the diameter of the other. What is the ratio of the moment of inertia of the large coin compared to the small coin? Take the axis of rotation to be perpendicular to the coin and through its center; assume that the coins have the same thickness. Hint: The moment of inertia of a solid disk about its center is \(\frac{1}{2} M R^{2}\).
Two forces produce equal torques on a door about the door hinge. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force has a greater magnitude?
Wheels \( A \) and \( B \) are connected by a moving belt and are both free to rotate about their centers. The belt does not slip on the wheels. The radius of Wheel \( B \) is twice the radius of Wheel \( A \). Wheel \( A \) has constant angular speed \( \omega_A \) and Wheel \( B \) has constant angular speed \( \omega_B \). Which of the following correctly relates \( \omega_A \) and \( \omega_B \)?
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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] | |
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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