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A particle moves in a circle of radius 1 m with initial angular speed 12 rad/s. At t = 0 it experiences a constant angular acceleration alpha and its angular speed rises to (480/pi) revolutions per minute in 2 s, after which it continues at that speed. Find (i) the angular acceleration, (ii) the tangential speed as a function of time, (iii) the magnitude of acceleration at t = 0.5 s and at t = 3 s, and (iv) the angular displacement at t = 3 s.

1. (i) 4 rad/s² (ii) v = 12 + 4t m/s (iii) a(0.5 s) = 60 m/s², a(3 s) = 156 m/s² (iv) 48 rad
2. (i) 4 rad/s² (ii) v = 12 + 4t m/s (iii) a(0.5 s) = 20 m/s², a(3 s) = 36 m/s² (iv) 24 rad
3. (i) 2 rad/s² (ii) v = 12 + 2t m/s (iii) a(0.5 s) = 30 m/s², a(3 s) = 78 m/s² (iv) 36 rad
4. (i) 8 rad/s² (ii) v = 12 + 8t m/s (iii) a(0.5 s) = 100 m/s², a(3 s) = 204 m/s² (iv) 72 rad

Correct answer: (i) 2 rad/s² (ii) v = 12 + 2t m/s (iii) a(0.5 s) = 30 m/s², a(3 s) = 78 m/s² (iv) 36 rad

Solution

Final angular speed = 16 rad/s, so alpha = (16 - 12)/2 = 2 rad/s² and v = 12 + 2t; the consistent option is the one with alpha = 2 rad/s² and v = 12 + 2t.

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