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 .

Centripetal Acceleration = 2.701 m/s²

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

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j