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| 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. |
Just ask: "Help me solve this problem."
A discus is held at the end of an arm that starts at rest. The average angular acceleration of [katex]54 \, \text{rad/s}^2 [/katex] 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.
Flywheels (rapidly rotating disks) are widely used in industry for storing energy. They are spun up slowly when extra energy is available, then decelerate quickly when needed to supply a boost of energy. A flywheel, \( 20 \, \text{cm}\) in diameter can spin at \( 20 \, \text{rpm}\).
The speed of a 40 N hockey puck, sliding across a level ice surface, decreases at the rate of 0.61 m/s2. The coefficient of kinetic friction between the puck and ice is
Three forces of equal magnitude are applied to a \( 3 \)-m by \( 2 \)-m rectangle. Force \( F_1 \) and \( F_2 \) act at \( 45^\circ \) angles to the vertical as shown, while \( F_3 \) acts horizontally.
Two uniform solid balls, one of radius \( R \) and mass \( M \), the other of radius \( 2R \) and mass \( 8M \), roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first?
\(\mu = \frac{2}{7}\tan\theta\)\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] |
General Metric Conversion Chart
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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