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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 Christmas ornament made from a thin hollow glass sphere hangs from a thin wire of negligible mass. It is observed to oscillates with a frequency of \( 2.50 \) \( \text{Hz} \) in a city where \( g = 9.80 \) \( \text{m/s}^2 \). What is the radius of the ornament? The moment of inertia of the ornament is given by \( I = \frac{5}{3} mr^2 \).
A construction worker spins a square sheet of metal of mass 0.040 kg with an angular acceleration of 10.0 rad/s2 on a vertical spindle (pin). What are the dimensions of the sheet if the net torque on the sheet is 1.00 N·m? Assume that the moment of inertia of a rectangle is [katex] I = \frac{1}{12}M(a^2+b^2) [/katex]
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 \( 20 \, \text{rpm}\).
A system consists of two small disks, of masses \( m \) and \( 2m \), attached to ends of a rod of negligible mass of length \( 3x \). The rod is free to turn about a vertical axis through point \( P \). The first mass, \( m \), is located \( x \) away from point \( P \), and therefore the other mass, of \( 2m \), is \( 2x \) from point \( P \). The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is \( \mu \). At time \( t = 0 \), the rod has an initial counterclockwise angular velocity \( \omega_i \) about \( P \). The system is gradually brought to rest by friction.
Derive expressions for the following quantities in terms of \( \mu \), \( m \), \( x \), \( g \), and \( \omega_i \).
A horizontal uniform meter stick of mass 0.2 kg is supported at its midpoint by a pivot point. A mass of 0.1 kg is attached to the left end of the meter stick, and another mass of 0.15 kg is attached to the right end of the meter stick. The meter stick is free to rotate in the horizontal plane around the pivot point. What is the tension in the string supporting the left end of the meter stick?
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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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