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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 solid ball of mass \( M \) and radius \( R \) has rotational inertia \( \frac{2}{5} M R^{2} \) about its center. It rolls without slipping along a level surface at speed \( v \) just before it begins rolling up an inclined plane. Which of the following expressions correctly represents the maximum vertical height the solid ball can ascend to when it rolls up the incline without slipping?
A ice skater that is spinning in circles has an initial rotational inertia Ii. You can approximate her shape to be a cylinder. She is spinning with velocity ωi. As she extends her arms she her rotational inertia changes by a factor of x and her angular velocity changes by a factor of y. Which one of the following options best describe x and y.
A sphere starts from rest and rolls down an incline of height \( H = 1.0 \) \( \text{m} \) at an angle of \( 25^\circ \) with the horizontal, as shown above. The radius of the sphere \( R = 15 \) \( \text{cm} \), and its mass \( m = 1.0 \) \( \text{kg} \). The moment of inertia for a sphere is \( \frac{2}{5}mR^2 \). What is the speed of the sphere when it reaches the bottom of the plane?
A boy and a girl are balanced on a massless seesaw. The boy has a mass of \(60 \, \text{kg}\) and the girl’s mass is \(50 \, \text{kg}\). If the boy sits \(1.5 \, \text{m}\) from the pivot point on one side of the seesaw, where must the girl sit on the other side for equilibrium?
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?
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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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