Answer:
(A) 570 rad
(B) 10 s
(C) 12.5 rad/s²
Explanation:
The equations of motion for circular motions are used.
(A)
At t = 2.00 s, the angular displacement, θ, is given by
[tex]\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}[/tex]
After this time, it decelerates through an angular displacement of 440 rad.
Total angular displacement = 130 + 440 rad = 570 rad
(B)
At the time the circuit breaker tips, the angular velocity is given by
[tex]\omega = \omega_0+\alpha t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}[/tex]
This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:
[tex]\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t[/tex]
Here, [tex]\theta_2=440[/tex] (the angular displacement during deceleration)
The subscripts, i and f, on ω denote the initial and final angular velocities during deceleration.
[tex]\omega_i = 100[/tex]
[tex]\omega_f = 0[/tex]
[tex]t = \dfrac{2\theta_2}{\omega_i} = \dfrac{2\times400}{100} = 8\ \text{s}[/tex]
This is the time for deceleration. The deceleration began at t = 2 s.
Hence, the wheel stops at t = 2 + 8 = 10 s.
(C)
The deceleration is given by
[tex]\alpha_R = \dfrac{\omega_f-\omega_i}{t} = \dfrac{0-100}{8} = -12.5\text{ rad/s}^2[/tex]
The negative sign appears because it is a deceleration.
