Question

The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor reverse direction? Express your answer to two significant figures and include the appropriate units. tt = nothing nothing Request Answer Part B Through what angle does the motor turn between t =0 s and the instant at which it reverses direction? Express your answer to two significant figures and include the appropriate units. ΔθΔ θ = nothing nothing Request Answer

Answer 1

Answer:

Explanation:

Given

$\omega =13-\frac{1}{2}\cdot t^2$

Motor reverse its direction when \omega =0

$13-0.5t^2=0$

$26=t^2$

$t=\sqrt{26}=5.099\approx 5.1 s$

(b)

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

$\int d\theta =\int_{0}^{5.1}\omega dt$

$\int d\theta =\int_{0}^{t}(13-.05t^2)dt$

$\theta =(13t-0.1667\times t^3)_0^{5.1}$

$\theta =44.192^{\circ}$