What is the centripetal acceleration of the bug?

Question

You start an old record player and notice a bug on the surface close to the edge of the record. The record has a diameter of 30 cm and completes 78 revolutions each minute. What is the centripetal acceleration of the bug?

Answer 1

Given

r = 15cm = 0.15m

N = 78 rev/min

Find

an = ?

Solution

f = 78/60 = 1.3 rev/sec

w = 2 * pi * f =2 * pi * 1.3 = 8.168 rad/s

V = w * r = 8.168 * 0.15 = 1.225 m/s

an = V^2 / r = 1.225^2/0.15 = 10 m/s^2

Alternatively

an = w^2 * r = (8.168)^2 * 0.15 = 10 m/s^2