Answer:
(a) 0.942 m
(b) 18.84 m/s
(c) 2366.3 m/s²
(d) 0.05 s
Explanation:
(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.
(b) Its frequency, f, is 1200 rev/min = [tex]\dfrac{1200}{60}[/tex]rev/s = 20 rev/s.
Its angular frequency, ω = 2πf = 2π × 20 = 40π
The speed is given by
v = ωr = 40π × 0.15 = 6π = 18.84 m/s
(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²
(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.
[tex]T = \dfrac{1}{f}[/tex]
T = 1/20 = 0.05 s
(a) the distance in one revolution should be 0.942 m
(b) The speed is 18.84 m/s
(c) The magnitude of its acceleration is 2366.3 m/s²
(d) The period of the motion is 0.05 s
(a) In one revolution, it travels via one circumference,
= 2πr
= 2 × 3.14 × 0.15 m
= 0.942 m.
(b) Since Its frequency, f, is 1200 rev/min
= rev/s
= 1200 / 60
= 20 rev/s.
Now
Its angular frequency,
ω = 2πf
= 2π × 20
= 40π
Now
The speed is
v = ωr
= 40π × 0.15
= 6π
= 18.84 m/s
(c) Its acceleration is provided by,
a = ω²r
= (40π)² × 0.15
= 2366.3 m/s²
(d) The period should be the inverse of the frequency since it is the time taken to finish one revolution.
So,
T = 1/20 = 0.05 s
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