# Part (a) – To determine the acceleration of the system
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | T_1 – T_2 = I \alpha | This equation represents the relationship between the tensions on either side of the pulley (T_1 and T_2), the moment of inertia of the pulley (I), and the angular acceleration (\alpha). |
| 2 | I = \frac{1}{2} m_p r^2 | The moment of inertia (I) of a disk-shaped pulley, where m_p is the mass of the pulley and r is its radius. |
| 3 | \alpha = a/r | The angular acceleration (\alpha) is related to the linear acceleration (a) of the falling mass by the radius of the pulley (r). |
| 4 | f_k = \mu_k T_s | The frictional force f_k is calculated using the coefficient of kinetic friction (\mu_k) and the tangential component of the tension T_s, which can be approximated as T_s = (T_1 + T_2)/2. |
| 5 | T_1 – T_2 – f_k r = I \alpha | 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 | (m_2 g – T_2) – (T_1 – m_1 g) = (m_1 + m_2) a | The forces acting on the blocks (m_1 g – T_1 downwards on m_1 and m_2 g – T_2 downwards on m_2, assuming m_1 < m_2 so m_2 descends) produces the net force, which equals the total system mass times the system’s acceleration. |
| 7 | Solve for a using the equations | Combine all equations replacing \alpha with a/r and solve for a 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 | m_1 g – T_1 = m_1 a | For mass m_1, the net force is the difference between the weight and the tension, which equals its mass times acceleration. |
| 2 | T_1 = m_1 g – m_1 a | Solving the above equation for T_1 gives the tension in the rope on the side of mass m_1. |
| 3 | T_1 calculation | Plug the values of g, m_1, and a obtained from part (a) to find T_1. |
# Part (c) – To determine the magnitude and direction of the frictional force exerted on the pulley
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | f_k = \mu_k \frac{T_1 + T_2}{2} | Using the approximation that the mean tension provides the force of friction calculation against the direction of motion. |
| 2 | f_k calculation | Calculate the value by substituting \mu_k, T_1, and T_2 values (where T_2 can be similarly calculated like T_1). |
| 3 | f_k 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 m_2 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 g = 9.8 \, \text{m/s}^2 ).
Phy can also check your working. Just snap a picture!
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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_1m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu N |
| R = \frac{v_i^2 \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
| a_c = \frac{v^2}{r} | PE = mgh |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m v | \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 |
|---|
| F = -k x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
| M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
| M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kgm/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (Nm)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
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