Question

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 2.91 s. Find the bug’s centripetal acceleration 1.5 s after the bug starts from rest. (1 in = 2.54 cm). Answer in units of m/s 2 .

Answer 1

Centripetal Acceleration = 2.701 m/s²

Step-by-step explanation:

Given:

Starting from rest so initial angular velocity ωi = 0 rad/s

Final angular velocity ωf= 78 rev/min =78 × 2π /60 rad/s = 8.168 rad/s

Radius of Rim =d/2 =12/6= 6 in = 6× 2.54 cm =15.24 cm = 0.1524 m

angular acceleration α = (ωf - ωi)/t = (8.168 rad/s - 0 rad/s)/2.91 s

α =2.807 rad/s²

now to find angular velocity at 1.5 s we have

ω = α × t = 2.807 rad/s² × 1.5 s = 4.210 rad/s

so centripetal acceleration ac = ω² r = (4.210 rad/s)² × 0.1524 m

ac= 2.701 m/s²