(a) Calculate the linear speed of the sphere when it reaches the bottom of the incline.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(h = L \sin(\theta)\) | Calculate the vertical height \(h\) fallen by the sphere using the length of the incline \(L\) and the sine of the incline angle \(\theta\). |
| 2 | \(h = 7.0 \sin(35^\circ)\) | Substitute \(L = 7.0 \, m\) and \(\theta = 35^\circ\). |
| 3 | \(PE_{\text{top}} = KE_{\text{trans}} + KE_{\text{rot}}\) | Use the conservation of mechanical energy, where potential energy at the top is equal to the sum of transnational kinetic energy and rotational kinetic energy at the bottom. |
| 4 | \(mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\) | Express the conservation of energy equation in terms of \(v\) (linear velocity) and \(\omega\) (angular velocity). |
| 5 | \(I = \frac{2}{5}MR^2\) | Substitute the given moment of inertia for a solid sphere, where \(I = \frac{2}{5}MR^2\). |
| 6 | \(v = R\omega\) | Relation between linear velocity and angular velocity for rolling without slipping. |
| 7 | \(\omega = \frac{v}{R}\) | Rearrange the equation for \(\omega\). |
| 8 | \(mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2\) | Substitute \(I\) and \(\omega\) in terms of \(v\) and \(R\). |
| 9 | \(mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2\) | Simplify the equation by canceling \(M\) and \(R\). |
| 10 | \(mgh = \frac{7}{10}mv^2\) | Combine like terms. |
| 11 | \(v^2 = \frac{10}{7}gh\) | Isolate \(v^2\). |
| 12 | \(v = \sqrt{\frac{10}{7}gh}\) | Take the square root to find \(v\). |
| 13 | \(v = \sqrt{\frac{10}{7}(9.8)(7.0 \sin(35^\circ))}\) | Substitute the values of \(g\) and \(h\). |
| 14 | \(\boxed{v \approx 7.5 \, \text{m/s}}\) | Final answer. |
(b) Determine the angular speed of the sphere at the bottom of the incline.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\omega = \frac{v}{R}\) | Use the relation between linear and angular velocities for a rolling object. |
| 2 | \(\omega = \frac{7.5}{0.15}\) | Substitute \(v = 7.5 \, \text{m/s}\) and \(R = 0.15 \, m\) (converted from cm). |
| 3 | \(\boxed{\omega \approx 50 \, \text{rad/s}}\) | Final answer. |
(c) Does the linear speed depend on the radius or mass of the sphere?
| Step | Analysis | Conclusion |
|---|---|---|
| 1 | From the energy conservation equation, the mass \(m\) cancelled out and the final expression for \(v\) didn’t include the radius \(R\). | The linear speed does not depend on the mass or radius of the sphere as both factors were eliminated in deriving \(v\). |
(d) Does the angular speed depend on the radius or mass of the sphere?
| Step | Analysis | Conclusion |
|---|---|---|
| 1 | Angular speed \(\omega\) was found from \(v\) divided by \(R\), but it did not involve mass \(m\). | Angular speed does depend on the radius and does not depend on the mass of the sphere. |
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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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