angular velocityangular acceleration

0901 Angular Velocity And Acceleration

  • Circumference: C=2πr=πdC = 2\pi r = \pi d

  • Arc length: S=rθS = r\th (θ\th has the unit of radian)

  • Period of Oscillation TT: Time to one complete full Revolution/Oscillation.

  • Frequency ff (Hz): number of revolution/scillation occur in a given time.

  • Angular Displacement: Δθ=Δθf−Δθi\Delta \th = \Delta \th_f - \Delta \th_i

  • Angular Velocity ω(rad/s)\omega (rad/s) , change in θ\th over time. ωavg=ΔθΔt\omega_{avg} = \dfrac{\Delta \th}{\Delta t}

  • Angular Acceleration a(rad/s2)a (rad/s^2). aavg=ωΔta_{avg} = \dfrac{\omega}{\Delta t}

  • Centripetal acceleration a⃗c(m/s2)\vec a_c (m/s^2). a⃗c=v2r\vec a_c = \dfrac{v^2}{r}

  • Tangential acceleration a⃗t(m/s2)\vec a_t (m/s^2)
    Object undergoing circular motion that is Non-uniform (speed changes).

  • Linear ⟷\longleftrightarrow Rotational Relations in Motion

v=rωa⃗c=v2r=rω2ω=2πf=2πT\begin{aligned} v &= r\omega \\ \vec a_c &= \frac{v^2}{r} = r\omega^2\\ \omega &= 2\pi f = \frac{2\pi}{T} \end{aligned}

Exerices

(9.2) An airplane propeller is rotating at 1980 rev/min.
a. Compute the propeller's angular velocity in rad/s.
b. How long in seconds does it take for the propeller to turn through 37°37 \degree?

Solution

a. ω=207rad/s\omega = 207 rad/s
b. t=3.11×10−3st = 3.11 \times 10^{-3}s

(9.14) A circular saw blade of diameter 0.205m0.205 m starts from rest. In a time interval of 6.35s6.35 s it accelerates with constant angular acceleration to an angular velocity of 133rad/s133 rad/s. Find the angular acceleration and the angle through which the blade has turned.

Solution

α=ωft=1336.35=20.9rad/s2θ=12(ω0+ωf)t=12(0+133)⋅6.35=422.3rad\begin{aligned} \alpha &= \frac{\omega_f}{t}\\ &= \frac{133}{6.35} = 20.9 rad/s^2\\ \th &= \frac{1}{2}(\omega_0 + \omega_f) t\\ &= \frac{1}{2}(0 + 133) \cdot 6.35 = 422.3 rad \end{aligned}

(9.16) At t=0t=0 a grinding wheel has an angular velocity of 29.0rad/s29.0 rad/s. It has a constant angular acceleration of 35.0rad/s235.0 rad/s^2 until a circuit breaker trips at time t=1.90st = 1.90s. From then on, it turns through an angle 431rad431 rad as it coasts to a stop at constant angular acceleration.
a. Through what total angle did the wheel turn between t=0t=0 and the time it stopped?
b. At what time did it stop?
c. What was its acceleration as it slowed down?

Solution

a. Let the angular displacement from t=0t=0 to t=1.9t=1.9 be θ1\th_1, and that from t=1.9t=1.9 to stop be θ2=431rad\th_2 = 431rad. The total angular displacement θ\th is

θ=θ1+θ2=ωt+12αt2+θ2=29.0⋅1.9+12⋅35.0⋅1.92+431=549rad\begin{aligned} \th &= \th_1 + \th_2\\ &= \omega t + \frac{1}{2} \alpha t^2 + \th_2\\ &= 29.0 \cdot 1.9 + \frac{1}{2} \cdot 35.0 \cdot 1.9^2 + 431\\ &= 549rad \end{aligned}

b. Let the angular acceleration after the breakers trips be α2\alpha_2

ωf2=ω02+2α2θ2ωf=0rad/sω0=29.0+35.0⋅1.9=1044rad/s⇒α2=−ω022θ2=−10.6rad/s2\begin{aligned} \omega_f^2 &= \omega_0^2 + 2\alpha_2 \th_2\\ \omega_f &= 0 rad/s\\ \omega_0 &= 29.0 + 35.0 \cdot 1.9 = 1044 rad/s\\ \To \alpha_2 &= -\frac{\omega_0^2}{2\th_2}\\ &= - 10.6 rad/s^2 \end{aligned}

c. t2=0−ω0α=9.0st_2 = 0 - \dfrac{\omega_0}{\alpha} = 9.0s. The total time is t=t1+t2=1.9+9=10.9st = t_1 + t_2 = 1.9 + 9 = 10.9s