| Step | Derivation/Formula | Reasoning |
|---|---|---|
| A1 | T – Mg = \frac{Mv^2}{L} | This is the force equilibrium equation at the bottom point. T is the tension in the string, M is the mass of the ball, g is the acceleration due to gravity, v is the velocity of the ball, and L is the length of the string. |
| A2 | T = 3Mg | The tension at the bottom is given to be three times the weight of the ball. |
| A3 | 3Mg – Mg = \frac{Mv^2}{L} | Substituting the tension value into the equilibrium equation. |
| A4 | v = \sqrt{2gL} | Simplifying the equation for v , observe M cancels out. This is velocity at any given point around the circle. |
| A5 | F_{\text{centripetal}} = \frac{Mv^2}{L} | At the top, we can use the velocity calculated in the previous step to find the centripetal force required to keep the ball moving in the circle. |
| A6 | 2mg | Substitute in the equation for velocity (from step A4) so that the final equation is in terms of M \, g \, L |
| B1 | v_{\text{top}} = \sqrt{v^2 – 4gL} | Using conservation of energy. KE_{\text{bottom}} + PE_{\text{bottom}} = KE_{\text{top}} + PE_{\text{top}} . The velocity at the top is found by noting the potential energy difference between top and bottom. Simplify by substituting v from A4. |
| B2 | v_{\text{top}} = \sqrt{2gL – 4gL} | = \sqrt{-2gL} \rightarrow which is zero since 2gL > 4gL |
| C1 | t = \sqrt{\frac{4L}{g}} | Ball falls freely under gravity and has no initial vertical velocity, so \delta y = \frac{1}{2}gt^2 ; solving for t gives the time to fall a distance L . Note that the displacement from the top to the bottom is twice the radius of the circle or 2L . |
| D1 | \Delta x = v_0t | The horiztontal distance traveled by any projectile is the product of the horiztonal speed and the time in air. |
| D2 | \sqrt{2gL} \times \sqrt{\frac{4L}{g}} | Substitute in velocity from Step A4 and time from step C1 |
| D3 | \sqrt{8}L | Simplify |
(a) The net force on the ball at the top is 2Mg , downward.
(b) The velocity of the ball at the top is v = \sqrt{2gL} .
(c) The time it takes to reach the ground is \sqrt{\frac{4L}{g}} .
(d) The horizontal distance traveled is \sqrt{8}L
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An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50.0° with respect to the horizontal. When the plane’s altitude is 732 m, the pilot releases a package.
A 2.2 kg ball on the end of a 0.35 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 5.3 m/s. Find the tension in the string.
A 2.2 \times 10^{21} \, \text{kg} moon orbits a distant planet in a circular orbit of radius 1.5 \times 10^8 \, \text{m}. It experiences a 1.1 \times 10^{19} \, \text{N} gravitational pull from the planet. What is the moon’s orbital period in earth days?
Two wires are tied to the 500 g sphere shown below. The sphere revolves in a horizontal circle at a constant speed of 7.2 m/s. What is the tension in the upper wire? What is the tension in the lower wire?
You kick a soccer ball with an initial velocity directed 53° above the horizontal. The ball lands on a roof 7.2 m high. The wall of the building is 25 m away, and it takes the ball 2.1 seconds to pass directly over the wall.
(a) 2Mg , downward.
(b) v = \sqrt{2gL} .
(c) \sqrt{\frac{4L}{g}} .
(d) \sqrt{8}L
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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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