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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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While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
How long does it take for a rotating object to speed up from 15.0 rad/s to 33.3 rad/s if it has a uniform angular acceleration of 3.45 rad/s2?
A massless rigid rod of length [katex]3d[/katex] is pivoted at a fixed point [katex]W[/katex], and two forces each of magnitude [katex]F[/katex] are applied vertically upward as shown above. A third vertical force of magnitude [katex]F[/katex] may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude [katex]F[/katex] is applied at point?
Three blocks of masses m3 = 1.0, m2 = 2.0, and m1 = 4.0 kilograms are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above.
The figure above shows a uniform beam of length L and mass M that hangs horizontally and is attached to a vertical wall. A block of mass M is suspended from the far end of the beam by a cable. A support cable runs from the wall to the outer edge of the beam. Both cables are of negligible mass. The wall exerts a force [katex]F_w[/katex] on the left end of the beam. For which of the following actions is the magnitude of the vertical component of [katex]F_w[/katex] smallest?
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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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