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3 years ago

9

A bug is on the rim of a 78 rev/min, 12 in. diameter record. The record moves from rest to its final angular speed in 2.67 s. Fi

nd the bug’s centripetal acceleration 1.5 s after the bug starts from rest. (1 in = 2.54 cm).

1 answer:

viva [34]3 years ago

3 0

Answer:

6.41m/s2

Explanation:

unit conversion:

78 rev/min = 78rev/min * 2π rad/rev * 1/60 min/sec = 8.168 rad/s

12 in = 12 in * 2.54cm/in = 30.48 cm = 0.3048 m

So the record accelerate from rest to 8.168 rad/s, its (assuming constant) angular acceleration would be

$\alpha = \Delta \omega / t = (8.168 - 0) / 2.67 = 3.06 rad/s^2$

From here we can calculate the angular velocity of the bug at t2 = 1.5s

$\omega_2= \alpha*t_2 = 3.06 * 1.5 = 4.59 rad/s$

Its centripetal acceleration would then be

$a = \omega_2^2*R = 4.59^2*0.3048 = 6.41 m/s^2$

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2 years ago

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