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A conical pendulum is formed by attaching a ball of mass m to a string of length L, then allowing the ball to move in a horizontal circle of radius R.
Finding tension in the string
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | \tan(\theta) = \frac{R}{h} | The angle \theta is between the string and the vertical axis. h is the vertical distance from the ball to the pivot. |
| 2 | h = L \cos(\theta) | From the geometry of the cone. |
| 3 | \tan(\theta) = \frac{R}{L \cos(\theta)} | Substituting the expression for h. |
| 4 | \sin(\theta) = \frac{R}{\sqrt{R^2 + h^2}} | From the right triangle formed by the string, h, and R. |
| 5 | F_{T} \sin(\theta) = F_{c} | The horizontal component of tension provides the centripetal force (F_{c}). |
| 6 | F_{c} = m \frac{v^2}{R} | Centripetal force formula. v is the velocity of the ball. |
| 7 | F_{T} \cos(\theta) = mg | The vertical component of tension balances gravity. |
| 8 | F_{T} = \frac{mg}{\cos(\theta)} | Isolating F_{T} in the vertical balance. |
| 9 | F_{T} = \frac{mg}{\cos(\theta)} = \frac{mg}{\sqrt{1 – \sin^2(\theta)}} | Using \cos(\theta) = \sqrt{1 – \sin^2(\theta)}. |
| 10 | F_{T} = \frac{mg}{\sqrt{1 – \left(\frac{R}{\sqrt{R^2 + h^2}}\right)^2}} | Substituting \sin(\theta). |
| 11 | \boxed{F_{T} = \frac{mg}{\sqrt{1 – \frac{R^2}{R^2 + L^2 \cos^2(\theta)}}}} | Final expression for tension, substituting h = L \cos(\theta). |
Finding period of the pendulum
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | F_{c} = m \frac{v^2}{R} | Centripetal force formula. |
| 2 | F_{T} \sin(\theta) = m \frac{v^2}{R} | The horizontal component of tension provides the centripetal force. |
| 3 | v = R \omega | Relationship between linear velocity and angular velocity (\omega). |
| 4 | m \frac{(R \omega)^2}{R} = F_{T} \sin(\theta) | Substituting v with R \omega. |
| 5 | \omega^2 = \frac{F_{T} \sin(\theta)}{mR} | Isolating \omega^2. |
| 6 | \omega = \sqrt{\frac{g}{R \tan(\theta)}} | Using F_{T} \sin(\theta) = mg \sin(\theta) and simplifying. |
| 7 | T = \frac{2\pi}{\omega} | Period (T |
A ball of mass m is fastened to a string. The ball swings at constant speed in a vertical circle of radius R with the other end of the string held fixed. Neglecting air resistance, what is the difference between the string’s tension at the bottom of the circle and at the top of the circle?
Two satellites are in circular orbits around Earth. Satellite A has speed vA . Satellite B has an orbital radius nine times that of satellite A. What is the speed of satellite B?
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 person’s back is against the inner wall of spinning cylinder with no support under their feet. If the radius is R, find an expression for the minimum angular speed so the person does not slide down the wall. The coefficient of static friction is µs.
Note: If you haven’t studied angular velocity \omega yet, just find the linear velocity v.
The distance from earth to sun is 1.0 AU. The distance from Saturn to sun is 9 AU. Find the period of Saturn’s orbit in years. You can assume that the orbits are circular.
Find the escape speed from a planet of mass 6.89 x 1025 kg and radius 6.2 x 106 m.
A block starts at rest on a frictionless inclined track which then turns into a circular loop of radius R and is vertical. In terms of R and constants, find the minimum height h above the bottom of the loop the block must start from so it makes it around the loop.
A satellite circling Earth completes each orbit in 132 minutes.
A car is safely negotiating an unbanked circular turn at a speed of 17 m/s on dry road. However, a long wet patch in the road appears and decreases the maximum static frictional force to one-fifth of its dry-road value. If the car is to continue safely around the curve, by what factor would the it need to change the original velocity?
An airplane can safely bank when subjected to a centripetal acceleration of 8 g’s. If the airplane flies at a constant speed of 400 m/s, how long does it take to make a 180° turn?
An Olympic bobsled team goes through a horizontal curve at a speed of 120 km/hr. If the radius of curvature is 10.0 m, what is the apparent weight the crew experiences-express in terms of mg.
What is a man’s apparent weight at the equator if his weight is 500 N? The earth’s radius is 6.37 x 106 m.
A rock is whirled on the end of a string in a horizontal circle of radius R with a constant period T. If the radius of the circle is reduced to R/3, while the period remains T, what happens to the centripetal acceleration (ac) of the rock?
Two identical object rests on a platform rotating at constant speed. Object A is at distance of half the platform’s radius from the center. Object B lays at edge of the platform. Assuming the platform continues rotating at the same speed, how does the centripetal force of the two objects compare?
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?
The ultracentrifuge is an important tool for separating and analyzing proteins. Because of the enormous centripetal accelerations, the centrifuge must be carefully balanced, with each sample matched by a sample of identical mass on the opposite side. Any difference in the masses of opposing samples creates a net force on the shaft of the rotor, potentially leading to a catastrophic failure of the apparatus. Suppose a scientist makes a slight error in sample preparation and one sample has a mass 10 mg larger than the opposing sample.
If the samples are 12 cm from the axis of the rotor and the ultracentrifuge spins at 60000 rpm, what is the magnitude of the net force on the rotor due to the unbalanced samples?
A concrete highway curve of radius 60.0 m is banked at a 12.0 ° angle. What is the maximum speed with which a 1300 kg rubber-tired car can take this curve without sliding? (Take the static coefficient of friction of rubber on concrete to be 1.0).
A 1.00 kg mass is attached to a 0.800 m long string and spun in a vertical circle. The mass completes 2.00 revolution in 1.00 s.
A car is safely negotiating an unbanked circular turn at a speed of 21 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a wet patch in the road decreases the maximum static frictional force to one-third its dry road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?
A car travels at a constant speed around a circular track whose radius is 2.6 km. the car goes once around the track in 360 s. What is the magnitude of the centripetal acceleration of the car?
A speed skater goes around a turn that has a radius of 31 m. The skater has a speed of 14 m/s and experiences a centripetal force of 460 N. What is the mass of the skater?
A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has coefficients of static and kinetic friction of µs = 0.80 and µk = 0.50. The turntable slowly speeds up to 60 rpm. Does the coin slide off the turntable?
A 2.00 x102 g block on a 50.0 cm long string swings in a circle on a horizontal, frictionless table at 75.0 rpm. What is the speed of block? What is the tension in the string?
Two identical satellites are placed in orbit of two different planets. Satellite A orbits Mars, and Satellite B orbits Jupiter. The orbital speeds of each satellite are the same. Which satellite has a greater orbital radius?
The International Space Station has a mass of 4.2 x105 kg and orbits Earth at a distance of 4.0 x102 km above the surface. Earth has a radius of 6.37 x106 m, and mass of 5.97 x1024 kg. Calculate the following:
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| Kinematics | Forces |
|---|---|
| \Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2 | F = m \cdot a |
| v = v_i + a \cdot t | F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu \cdot N |
| R = \frac{v_i^2 \cdot \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{m \cdot v^2}{r} | KE = \frac{1}{2} m \cdot v^2 |
| a_c = \frac{v^2}{r} | PE = m \cdot g \cdot h |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m \cdot v | \tau = r \cdot F \cdot \sin(\theta) |
| J = \Delta p | I = \sum m \cdot r^2 |
| p_i = p_f | L = I \cdot \omega |
| Simple Harmonic Motion |
|---|
| F = -k \cdot 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} |
| 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 (kg·m/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (N·m)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kg·m}^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} |
