Angular Kinematics Study Notes

Angular Kinematics

• Using knowledge and skills to predict the behavior of point masses undergoing circular motion.
• Analogous to constant linear acceleration equations.
• Constant acceleration and constant velocity equations are of similar form in angular and linear conditions.

Equations

• Constant acceleration and constant velocity equations are of similar form in angular and linear conditions.

Example Problem: Discus Thrower

• Initial angular velocity ($\omega_i$) = 0 radians per second.
• Final angular velocity ($\omega_f$) = 15 radians per second.
• Radius (r) = 0.81 meters.
• Time interval ($\Delta t$) = 0.27 seconds.
• Goal: Find linear acceleration (magnitude and direction).
• Two components of linear acceleration: tangential and centripetal.
• Centripetal acceleration: $a_c = \frac{v^2}{r}$.
  • Linear Speed: $v = \omega * r$.
  • Centripetal acceleration becomes: $a_c = \omega^2 * r$.
  • $a_c = (15)^2 * 0.81 = 182$ meters per second squared.
• Tangential acceleration: $a_t = \alpha * r$.
  • $\alpha = \frac{\Delta \omega}{\Delta t}$.
  • $\alpha = \frac{15}{0.27} = 55.6$ radians per second squared.
  • $a_t = 55.6 * 0.81 = 45$ meters per second squared.
• Magnitude of linear acceleration: $a = \sqrt{a<em>c^2 + a</em>t^2}$.
  • $a = \sqrt{182^2 + 45^2} = 188$ meters per second squared.
• Direction: $\theta = tan^{-1}(\frac{a<em>t}{a</em>c})$.
  • $\theta = tan^{-1}(\frac{45}{182}) = 14$ degrees.

Example Problem: Automatic Dryer

• Initial angular velocity ($\omega_i$) = 0 radians per second.
• Final angular velocity ($\omega_f$) = 5.2 radians per second.
• Angular acceleration ($\alpha$) = 4 radians per second squared.
• Goal: Find the time it takes to reach operating speed ($\Delta t$).
• Equation: $\omega<em>f = \omega</em>i + \alpha * t$.
  • $t = \frac{\Delta \omega}{\alpha}$.
  • $t = \frac{5.2}{4} = 1.3$ seconds.

Example Problem: Human Centrifuge

• Angular velocity ($\omega$) = 2.5 radians per second.
• Centripetal acceleration ($a_c$) = 3.2g = 31 meters per second squared.
• Goal: Find the length of the centrifuge arm (r).
• $a_c = r * \omega^2$.
  • $r = \frac{a_c}{\omega^2}$.
  • $r = \frac{31}{(2.5)^2} = 5$ meters.

Example Problem: Human Centrifuge (Part 2)

• Centrifuge speeds up from rest to 2.5 radians per second with constant angular acceleration.
• Net linear acceleration magnitude = 4.8g = 47 meters per second squared.
• Centripetal acceleration ($a_c$) = 3.2g = 31 meters per second squared.
• Goal: Find the average angular acceleration ($\alpha$).
• Tangential acceleration: $a_t = r * \alpha$.
  • $\alpha = \frac{a_t}{r}$.
• Finding tangential acceleration using Pythagorean theorem:
  • $a<em>t = \sqrt{a^2 - a</em>c^2}$.
  • $a_t = \sqrt{47^2 - 31^2} = 35$ meters per second squared.
• $\alpha = \frac{35}{5} = 7$ radians per second squared.

Example Problem: House Fan

• Fan starts from rest and speeds up to 10 revolutions per second after two revolutions.
• Goal: Find the angular acceleration of the fan.
• 1 revolution = 360 degrees = $2\pi$ radians.
• 2 revolutions = 720 degrees = $4\pi$ radians.
• Equation: $\omega<em>f^2 - \omega</em>i^2 = 2 * \alpha * \Delta \theta$.
  • $\alpha = \frac{\omega<em>f^2 - \omega</em>i^2}{2 * \Delta \theta}$.
• $\alpha = \frac{10^2 - 0^2}{2 * 2} = 23$ revolutions per second squared.

Summary

• Angular acceleration equations are used to solve for unknown kinematic variables.
• Similar thought process to solving linear kinematic problems.
• Next lecture: Examples of rotational motion that apply to the human body.