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(a) Calculate the linear speed of the sphere when it reaches the bottom of the incline.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]h = L \sin(\theta)[/katex] | Calculate the vertical height [katex]h[/katex] fallen by the sphere using the length of the incline [katex]L[/katex] and the sine of the incline angle [katex]\theta[/katex]. |
| 2 | [katex]h = 7.0 \sin(35^\circ)[/katex] | Substitute [katex]L = 7.0 \, m[/katex] and [katex]\theta = 35^\circ[/katex]. |
| 3 | [katex]PE_{\text{top}} = KE_{\text{trans}} + KE_{\text{rot}}[/katex] | 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 | [katex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/katex] | Express the conservation of energy equation in terms of [katex]v[/katex] (linear velocity) and [katex]\omega[/katex] (angular velocity). |
| 5 | [katex]I = \frac{2}{5}MR^2[/katex] | Substitute the given moment of inertia for a solid sphere, where [katex]I = \frac{2}{5}MR^2[/katex]. |
| 6 | [katex]v = R\omega[/katex] | Relation between linear velocity and angular velocity for rolling without slipping. |
| 7 | [katex]\omega = \frac{v}{R}[/katex] | Rearrange the equation for [katex]\omega[/katex]. |
| 8 | [katex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2[/katex] | Substitute [katex]I[/katex] and [katex]\omega[/katex] in terms of [katex]v[/katex] and [katex]R[/katex]. |
| 9 | [katex]mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2[/katex] | Simplify the equation by canceling [katex]M[/katex] and [katex]R[/katex]. |
| 10 | [katex]mgh = \frac{7}{10}mv^2[/katex] | Combine like terms. |
| 11 | [katex]v^2 = \frac{10}{7}gh[/katex] | Isolate [katex]v^2[/katex]. |
| 12 | [katex]v = \sqrt{\frac{10}{7}gh}[/katex] | Take the square root to find [katex]v[/katex]. |
| 13 | [katex]v = \sqrt{\frac{10}{7}(9.8)(7.0 \sin(35^\circ))}[/katex] | Substitute the values of [katex]g[/katex] and [katex]h[/katex]. |
| 14 | [katex]\boxed{v \approx 7.5 \, \text{m/s}}[/katex] | Final answer. |
(b) Determine the angular speed of the sphere at the bottom of the incline.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]\omega = \frac{v}{R}[/katex] | Use the relation between linear and angular velocities for a rolling object. |
| 2 | [katex]\omega = \frac{7.5}{0.15}[/katex] | Substitute [katex]v = 7.5 \, \text{m/s}[/katex] and [katex]R = 0.15 \, m[/katex] (converted from cm). |
| 3 | [katex]\boxed{\omega \approx 50 \, \text{rad/s}}[/katex] | 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 [katex]m[/katex] cancelled out and the final expression for [katex]v[/katex] didn’t include the radius [katex]R[/katex]. | The linear speed does not depend on the mass or radius of the sphere as both factors were eliminated in deriving [katex]v[/katex]. |
(d) Does the angular speed depend on the radius or mass of the sphere?
| Step | Analysis | Conclusion |
|---|---|---|
| 1 | Angular speed [katex]\omega[/katex] was found from [katex]v[/katex] divided by [katex]R[/katex], but it did not involve mass [katex]m[/katex]. | Angular speed does depend on the radius and does not depend on the mass of the sphere. |
Just ask: "Help me solve this problem."
A uniform meter stick has a mass of \( 45.0 \) \( \text{g} \) placed at the \( 20 \) \( \text{cm} \) mark. If a pivot is placed at the \( 42.5 \) \( \text{cm} \) mark and the meter stick remains horizontal in static equilibrium, what is the mass of the meter stick?
Consider a rigid body that is rotating. Which of the following is an accurate statement?
In a demonstration, a teacher holds the axle of a wheel that is spinning with constant angular speed. The teacher then releases the axle and the wheel begins to fall toward the ground. As the wheel falls, its angular speed remains constant. Which of the following correctly describes how the rotational kinetic energy \( K_{\text{rot}} \) of the wheel and the total kinetic energy \( K_{\text{tot}} \) of the wheel change, if at all, after the wheel is released but before it reaches the ground?
| \( K_{\text{rot}} \) | \( K_{\text{tot}} \) | |
|---|---|---|
| A | Constant | Constant |
| B | Constant | Increasing |
| C | Increasing | Constant |
| D | Increasing | Increasing |
A uniform rod of mass \( M_0 \) and length \( L \) is free to rotate about a pivot at its left end and is released from rest when the rod is \( 30^{\circ} \) below the horizontal, as shown in the figure. With respect to the pivot, the rod has rotational inertia \( I_0 = \dfrac{1}{3} M_0 L^2 \). Which of the following expressions correctly represents the magnitude of the net torque exerted on the rod about the pivot at the moment the rod is released?
A high-speed drill rotating counterclockwise at \( 2400 \) \( \text{rpm} \) comes to a halt in \( 2.5 \) \( \text{s} \).
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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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