2) A deep-sea fisherman hooks a big fish that swims away from the boat pulling t

Question

2) A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fr reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 6 cm from its axis I is given an angular acceleration of 90 rad/s' for 3.00 s, calculate the following (a) What is the final angular velocity of the reel? (b) At what speed is fishing line leaving the reel after 3.00 s elapses? (c) How many revolutions does the reel make? (d) How many meters of fishing line come off the reel in this time?.

Explanation / Answer

2)Given,

r = 6 cm = 0.06 m ; alpha = 90 rad/s^2 ; t = 3 s

a)we know from eqn in circular motion

w = w0 + alpha t

w = 0 + 90 x 3 = 270 rad/s

Hence, w = 270 rad/s

b)v = r w

v = 0.06 x 270 = 16.2 m/s

Hence, v = 16.2 m/s

c)we know that

theta = w0t + 1/2 alpha t^2

theta= 0 + 0.5 x 90 x 3^2 = 405 rad

so the no of revolutions will be

n = 405/2 pi = 64.5 rev

Hence, n = 64.5 rev

d)s = r theta

s = 0.06 x 405 = 24.3 m

hence, s = 24.3 m

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