| Derivation or Formula | Reasoning |
|---|---|
| \[ Mg\sin\theta – f = Ma \] | This is Newton’s second law for translation along the incline, where \(Mg\sin\theta\) is the component of gravity down the plane and \(f\) is the friction force. |
| \[ fR = I\alpha \quad \text{with} \quad a = \alpha R \] | The friction force provides the torque needed for rolling without slipping, and the rolling condition relates linear acceleration \(a\) to angular acceleration \(\alpha\). |
| \[ f = \frac{Ia}{R^2} \] | Rearranging the torque equation yields an expression for \(f\) in terms of \(a\). |
| \[ f = \frac{2}{5}Ma \] | Substitute the moment of inertia for a uniform solid sphere \(I = \frac{2}{5}MR^2\) into the previous expression. |
| \[ Mg\sin\theta – \frac{2}{5}Ma = Ma \] | Insert \(f = \frac{2}{5}Ma\) into Newton’s second law and set up the equation for \(a\). |
| \[ Mg\sin\theta = \left(1+\frac{2}{5}\right)Ma = \frac{7}{5}Ma \] | Simplify the equation to combine like terms and isolate \(a\). |
| \[ a = \frac{5}{7}g\sin\theta \] | Solve for the linear acceleration \(a\) of the sphere along the incline. |
| \[ f = \frac{2}{5}M\left(\frac{5}{7}g\sin\theta\right) = \frac{2}{7}Mg\sin\theta \] | Determine the friction force required for rolling by substituting \(a\) back into \(f = \frac{2}{5}Ma\). |
| \[ \frac{2}{7}Mg\sin\theta = \mu Mg\cos\theta \] | For the sphere to roll without slipping, the required friction force must be available, i.e., it must equal the maximum static friction \(\mu Mg\cos\theta\). |
| \[ \mu = \frac{2}{7}\tan\theta \] | Solve for the minimum coefficient of friction \(\mu\) needed to prevent slipping. |
| \[ \boxed{\mu = \frac{2}{7}\tan\theta} \] | This boxed expression is the final answer for part (a). |
| Derivation or Formula | Reasoning |
|---|---|
| \[ Mgh = \frac{1}{2}Mv_x^2 + \frac{1}{2}I\left(\frac{v_x}{R}\right)^2 \quad \Rightarrow \quad v_x = \sqrt{\frac{10}{7}gh} \] | For a rolling sphere, the gravitational potential energy converts into both translational and rotational kinetic energy. |
| \[ Mgh = \frac{1}{2}Mv_x^2 \quad \Rightarrow \quad v_x = \sqrt{2gh} \] | Without friction (\(\mu = 0\)), the sphere slides without rotating, so all potential energy becomes translational kinetic energy. |
| \[ \sqrt{2gh} > \sqrt{\frac{10}{7}gh} \] | Since \(2 > \frac{10}{7}\), the translational speed of a sliding sphere is higher than that of a rolling sphere, where some energy goes into rotation. |
| \(\text{Speed is greater when } \mu = 0\) | Thus, with zero friction the sphere attains a higher speed at the bottom because no energy is diverted to rotational motion. |
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Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the second force is applied at \( 30^\circ \) to the plane of the door. Which force exerts the greater torque about the door hinge?
A wheel of moment of inertia of \( 5.00 \) \( \text{kg} \cdot \text{m}^2 \) starts from rest and accelerates under a constant torque of \( 3.00 \) \( \text{N} \cdot \text{m} \) for \( 8.0 \) \( \text{s} \). What is the wheel’s rotational kinetic energy at the end of \( 8.0 \) \( \text{s} \)?
A block starts from rest at the top of a \(50^\circ\) incline. The coefficient of kinetic friction between the block and the incline is \(0.4\). If the block reaches a velocity of \(7 \, \text{m/s}\) at the bottom of the incline, what is the length of the incline?
Angular momentum cannot be conserved if
A uniform ladder with mass \( m_2 \) and length \( L \) rests against a smooth wall. A do-it-yourself enthusiast of mass \( m_1 \) stands on the ladder a distance \( d \) from the bottom (measured along the ladder). The ladder makes an angle \( \theta \) with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude \( f \) between the floor and the ladder. \( N_1 \) is the magnitude of the normal force exerted by the wall on the ladder, and \( N_2 \) is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive.
A solid sphere is rotating about an axis through its center at a constant rotation rate. Another hollow sphere of the same mass and radius is rotating about its axis through the center at the same rotation rate. Which sphere has a greater rotational kinetic energy?
If a constant net torque is applied to an object it will (select all that applies):
A meterstick is supported at its center, which is aligned with the center of a cradle located at position \( x = 0 \) \( \text{m} \). Two identical objects of mass \( 1.0 \) \( \text{kg} \) are suspended from the meterstick. One object hangs \( 0.25 \) \( \text{m} \) to the left of the support point, and the other object hangs \( 0.50 \) \( \text{m} \) to the right of the support point. The system is released from rest and is free to rotate. Which of the following claims correctly describes the subsequent motion of the system containing the meterstick, cradle, and the two objects?
A coffee cup on the dashboard of a car slides forward when the driver decelerates from \(45 ~ \frac{\text{km}}{\text{hr}}\) to rest in \(3.5 \, \text{s}\) or less. What is the coefficient of static friction between the cup and the dash? Assume the road and the dashboard are completely horizontal.
Consider a rigid body that is rotating. Which of the following is an accurate statement?
\(\mu = \frac{2}{7}\tan\theta\)\n\(\text{Greater}\)
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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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