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# Part (a) – To determine the acceleration of the system
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]T_1 – T_2 = I \alpha [/katex] | This equation represents the relationship between the tensions on either side of the pulley ([katex]T_1[/katex] and [katex]T_2[/katex]), the moment of inertia of the pulley ([katex]I[/katex]), and the angular acceleration ([katex]\alpha[/katex]). |
| 2 | [katex]I = \frac{1}{2} m_p r^2[/katex] | The moment of inertia ([katex]I[/katex]) of a disk-shaped pulley, where [katex]m_p[/katex] is the mass of the pulley and [katex]r[/katex] is its radius. |
| 3 | [katex]\alpha = a/r[/katex] | The angular acceleration ([katex]\alpha[/katex]) is related to the linear acceleration ([katex]a[/katex]) of the falling mass by the radius of the pulley ([katex]r[/katex]). |
| 4 | [katex]f_k = \mu_k T_s[/katex] | The frictional force [katex]f_k[/katex] is calculated using the coefficient of kinetic friction ([katex]\mu_k[/katex]) and the tangential component of the tension [katex]T_s[/katex], which can be approximated as [katex]T_s = (T_1 + T_2)/2[/katex]. |
| 5 | [katex]T_1 – T_2 – f_k r = I \alpha[/katex] | Combine the effects of tension and frictional force (considering the friction opposes the motion thus negative), and include the effect of the pulley’s inertia. |
| 6 | [katex](m_2 g – T_2) – (T_1 – m_1 g) = (m_1 + m_2) a[/katex] | The forces acting on the blocks ([katex]m_1 g – T_1[/katex] downwards on [katex]m_1[/katex] and [katex]m_2 g – T_2[/katex] downwards on [katex]m_2[/katex], assuming [katex]m_1[/katex] < [katex]m_2[/katex] so [katex]m_2[/katex] descends) produces the net force, which equals the total system mass times the system’s acceleration. |
| 7 | Solve for [katex]a[/katex] using the equations | Combine all equations replacing [katex]\alpha[/katex] with [katex]a/r[/katex] and solve for [katex]a[/katex] to get the system’s acceleration, involving simplifications and algebraic manipulation. |
# Part (b) – To determine the tension in the rope
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]m_1 g – T_1 = m_1 a[/katex] | For mass [katex]m_1[/katex], the net force is the difference between the weight and the tension, which equals its mass times acceleration. |
| 2 | [katex]T_1 = m_1 g – m_1 a[/katex] | Solving the above equation for [katex]T_1[/katex] gives the tension in the rope on the side of mass [katex]m_1[/katex]. |
| 3 | [katex]T_1[/katex] calculation | Plug the values of [katex]g[/katex], [katex]m_1[/katex], and [katex]a[/katex] obtained from part (a) to find [katex]T_1[/katex]. |
# Part (c) – To determine the magnitude and direction of the frictional force exerted on the pulley
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]f_k = \mu_k \frac{T_1 + T_2}{2}[/katex] | Using the approximation that the mean tension provides the force of friction calculation against the direction of motion. |
| 2 | [katex]f_k[/katex] calculation | Calculate the value by substituting [katex]\mu_k[/katex], [katex]T_1[/katex], and [katex]T_2[/katex] values (where [katex]T_2[/katex] can be similarly calculated like [katex]T_1[/katex]). |
| 3 | [katex]f_k[/katex] direction determination | The frictional force opposes the direction of relative motion between the rope and the pulley, which will be clockwise due to the motion of [katex]m_2[/katex] falling downward. |
This sequence of tables will help in understanding the steps involved thoroughly to solve the given problem numerically. Please note the precise solutions would require the use of actual numerical values derived or given directly in the problem (e.g., gravitational acceleration [katex] g = 9.8 \, \text{m/s}^2 [/katex]).
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When a fan is turned off, its angular speed decreases from 10 rad/s to 6.3 rad/s in 5.0 s. What is the magnitude of the average angular acceleration of the fan?
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 . At time t = 0, the rod has an initial counterclockwise angular velocity ωi about P. The system is gradually brought to rest by friction.
Derive an expressions for the following quantities in terms of µ, m, x, g, and ωi.
A boy is sitting at a distance [katex] d_1 [/katex] from the fulcrum, and girl is sitting at a distance [katex] d_2 [/katex] from the fulcrum, with [katex] d_1 > d_2 [/katex]. The seesaw is level, with the two ends at the same height. Derive an equation for the minimum mass of the seesaw that will keep it balanced with the two children on it.
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 spinning ice skater on extremely smooth ice is able to control the rate at which she rotates by pulling in her arms. Which of the following statements are true about the skater during this process?
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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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