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# Part (a): Angular acceleration at [katex] t = 4.0 \, \text{s} [/katex]
Note – To solve without calculus, graph the given function. The slope of this angular velocity v. time graph is acceleration. Therefore, approximate the slope at 4 seconds to find angular acceleration.
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | [katex]\omega = 20 – \frac{1}{2} t^2 \, \text{rad/s} [/katex] | Given expression for the angular velocity of the gear as a function of time. |
2 | [katex]\alpha = \frac{d\omega}{dt} [/katex] | Angular acceleration is the rate of change of angular velocity with respect to time. |
3 | [katex]\alpha = \frac{d}{dt} \left( 20 – \frac{1}{2} t^2 \right) = – t \, \text{rad/s}^2 [/katex] | Differentiate the given expression for [katex]\omega[/katex] with respect to [katex]t[/katex]. |
4 | [katex]\alpha(t=4) = -4 \, \text{rad/s}^2 [/katex] | Substitute [katex]t = 4.0 \, \text{s}[/katex] to find the angular acceleration at that instant. |
5 | [katex]\boxed{-4 \, \text{rad/s}^2} [/katex] | Final boxed numerical answer for the angular acceleration at [katex]t = 4 \, \text{s}[/katex]. |
# Part (b): Tangential acceleration of a tooth on the gear at [katex] t = 4.0 \, \text{s} [/katex]
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | [katex]a_t = \alpha \cdot r[/katex] | Tangential acceleration is the product of the angular acceleration and the radius of the gear. |
2 | [katex]r = \frac{d}{2} = \frac{6.0 \, \text{cm}}{2} = 3.0 \, \text{cm} = 0.03 \, \text{m}[/katex] | Calculate the radius of the gear from the given diameter (convert cm to m). |
3 | [katex]a_t = -4 \, \text{rad/s}^2 \times 0.03 \, \text{m} = -0.12 \, \text{m/s}^2[/katex] | Use the angular acceleration found in part (a) and the radius to determine the tangential acceleration. |
4 | [katex]\boxed{-0.12 \, \text{m/s}^2}[/katex] | Final boxed numerical answer for the tangential acceleration of a tooth on the gear at [katex]t = 4 \, \text{s}[/katex]. |
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In lacrosse, a typical throw is made by rotating the stick through an angle of roughly 90°, then releasing the ball when the stick is vertical, as shown above. If the 1 meter long stick is at rest when horizontal and the ball leaves the stick with a velocity of 10 m/s, what angular acceleration must the stick experience?
A student is asked to design an experiment to determine the change in angular momentum of a disk that rotates about its center and the product of the average torque applied to the disk and the time interval in which the torque is exerted. A net force is applied tangentially to the surface of the disk. The rotational inertia of the disk about its center is [katex]I = MR^2[/katex]. Which two of the following quantities should the student measure to determine the change in angular momentum of the disk after 10 s? Select two answers.
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Kinematics | Forces |
---|---|
[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
[katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
[katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
Circular Motion | Energy |
---|---|
[katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
[katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
[katex]KE_i + PE_i = KE_f + PE_f[/katex] |
Momentum | Torque and Rotations |
---|---|
[katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
[katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
[katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
Simple Harmonic Motion |
---|
[katex]F = -k x[/katex] |
[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
Constant | Description |
---|---|
[katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
[katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
[katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
[katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
Variable | SI Unit |
---|---|
[katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
[katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
[katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
[katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
[katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
Variable | Derived SI Unit |
---|---|
[katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
[katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
[katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
[katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
[katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
[katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
[katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\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}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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