Step | Calculation/Analysis | Description |
---|---|---|
1 | T_{\text{max, stationary}} = m_{\text{operator}} \times g | Maximum stationary tension with the operator’s weight. m_{\text{operator}} = 90 , \text{kg}, g = 9.81 , \text{m/s}^2. |
2 | v = \frac{2\pi \times L}{T} | Rotation speed calculation. L = 10 , \text{m}, T = 3 , \text{s}. |
3.1 | T_{\text{horizontal}} = F_{\text{centripetal}} | Horizontal component of tension equals centripetal force. |
3.2 | F_{\text{centripetal}} = \frac{m_{\text{child}} \times v^2}{r} | Centripetal force formula. m_{\text{child}} = 25 , \text{kg}. |
3.3 | r = L \sin(\theta) | Effective radius related to the angle θ. |
3.4 | F_{\text{centripetal}} = \frac{m_{\text{child}} \times v^2}{L \sin(\theta)} | Substitute r in centripetal force formula. |
3.5 | T_{\text{vertical}} = m_{\text{child}} \times g | Vertical component of tension equals child’s weight. |
3.6 | \tan(\theta) = \frac{T_{\text{horizontal}}}{T_{\text{vertical}}} | Relation between tension components. |
3.7 | \tan(\theta) = \frac{m_{\text{child}} \times v^2 / (L \sin(\theta))}{m_{\text{child}} \times g} | Substitute velocity and radius in the tangent formula. |
3.8 | cos\theta = \frac{9}{4\pi^2} | Final, simplified equation |
3.9 | \theta = \approx 76.8 | Approximate value of the angle to the vertical of any rider (angle does not depend of mass). |
4 | T_{\text{ride}}cos\theta = mg | Solve for tension to get a total tension of ~ 1290 N for when the child sits in the seat. Remember the to add the mass of the seat to the child. |
5 | Safety Assessment | Tension in the chain is greater while child is rotating than when operator is sitting in it stationary, indicating the chain might break when the child is rotating in it. |
M
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A delivery truck is traveling north. It then turns along a leftward circular curve. This the packages in the truck to slide to the RIGHT. Which of the following is true of the net force on the packages as they are sliding?
Suppose you are a passenger traveling in car along a road that bends to the left. Why will you feel like you are being thrown against the door. What causes this force?
A child has a toy tied to the end of a string and whirls the toy at constant speed in a horizontal circular path of radius R. The toy completes each revolution of its motion in a time period T. What is the magnitude of the acceleration of the toy (in terms of T, R, and g)?
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.
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.
No it is not safe for the child. Based on the rotational speed, the child would be raised to an angle of ~77° from the initial vertical position. At this point the tension in the chain (~1290 N), while rotating, is considerable greater than the maximum force of tension (~920 N) the chain can with stand.
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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} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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