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Answer:
(A) The magnitude of its angular acceleration 1.09 x 10⁻³ rad/s²
(B) Time of motion 240.2 s
Explanation:
Given;
final angular speed, ωf = 2.5 RPM
angular distance, θ = 5 rev = 5 x 2π = 10π rad
initial angular speed, ωi = 0
final angular speed in rad/s;
[tex]\omega_f = \frac{2.5 \ rev}{min} \times \ \frac{2\pi}{1 \ rev} \times \ \frac{1 \min}{60 s} = 0.2618 \ rad/s \\[/tex]
(A) the magnitude of its angular acceleration(rad/s^2);
[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\(0.2618)^2 = 0 + (2\times 10\pi)\alpha\\\\0.0685 = 20\pi \alpha\\\\\alpha = \frac{0.0685}{20\pi} \\\\\alpha = 1.09\times 10^{-3} \ rad/s^2[/tex]
(B) Time of motion;
[tex]\omega_f = \omega_i + \alpha t\\\\0.2618 = 0 + 1.09\times 10^{-3} t\\\\t = \frac{0.2618}{1.09\times 10^{-3}} \\\\t = 240.2 \ s[/tex]