Before solving the question, we can find the radius of the ball using Pythagorean theorem to get .866 m. We can also use the trig to solve for the angle each rope makes with the horizontal (30° for both ropes).
Sum of Forces in the Horizontal Direction:
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | \cos(30) = \frac{\sqrt{3}}{2} | Cosine of 30^\circ. |
| 2 | F_{\text{centripetal}} = \frac{mv^2}{r} | Centripetal force for circular motion. |
| 3 | T_1 \cos(\theta) + T_2 \cos(\theta) = \frac{mv^2}{r} | Sum of horizontal components of tension equals centripetal force. |
| 4 | T_1 \frac{\sqrt{3}}{2} + T_2 \frac{\sqrt{3}}{2} = \frac{(0.5)(7.2)^2}{0.866} | Substitute values for m, v, r, and \cos(\theta). |
| 5 | \frac{\sqrt{3}}{2}(T_1 + T_2) = 29.93 | Calculate centripetal force and factor out \frac{\sqrt{3}}{2}. |
Sum of Forces in the Vertical Direction:
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | \sin(30) = \frac{1}{2} | Sine of 30^\circ. |
| 2 | w = mg | Weight of the sphere. |
| 3 | T_2 \sin(\theta) + mg – T_1 \sin(\theta) = 0 | Vertical forces must balance: upward tensions and downward weight. |
| 4 | T_2 \frac{1}{2} + (0.5)(9.8) – T_1 \frac{1}{2} = 0 | Substitute values for m, g, and \sin(\theta). |
| 5 | \frac{1}{2}(T_2 – T_1) + 4.9 = 0 | Factor out \frac{1}{2} and calculate weight. |
Solving for Tensions:
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | Solve equations | Use the system of equations to solve for T_1 and T_2. |
| 2 | T_1 \approx 22.18 \text{ N} | Numerical solution for T_1. |
| 3 | T_2 \approx 12.38 \text{ N} | Numerical solution for T_2. |
Final Tensions:
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A ball of mass M is attached to a string of length L. It moves in a vertical circle and at the bottom the ball just clears the ground. The tension at the bottom of the path is 3 times the weight of the ball. Give all answers in terms of M, L, and g.
Friction provides the force needed for a car to travel around a flat, circular race track. Answer the following:
A 2.0 kg ball on the end of a 0.65 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 4.0 m/s. Find the tension in the string.
A compressed spring mounted on a disk can project a small ball. When the disk is not rotating, as shown in the top view above, the ball moves radially outward. The disk then rotates in a counterclockwise direction as seen from above, and the ball is projected outward at the instant the disk is in the position shown above. Which of the following best shows the subsequent path of the ball relative to the ground?
A pendulum consists of a ball of mass m suspended at the end of a massless cord of length L . The pendulum is drawn aside through an angle of 60° with the vertical and released. At the low point of its swing, the speed of the pendulum ball is
Upper wire: 22 N; Lower wire: 12 N
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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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