Question

Magnitude of the tangential acceleration of a bug on the rim of a 12.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 79.0 rev/min in 4.30 s? m/s2 (b) When the disk is at its final speed, what is the magnitude of the tangential velocity of the bug? m/s (c) One second after the bug starts from rest, what is the magnitude of its tangential acceleration? m/s2

Answer 1

a). a=0.289 m/s²

b). v= 1.243 m/s

c).a=0.289 m/s²

Step-by-step explanation:

First let's convert to metric:

radius r = ½ * 12in * 2.54cm/in = 15.24 cm

and find the final angular velocity:

ω = 79rev/min * 2πrad/rev * 1min/60s = 8.16 rad/s

(a)

α = Δω / Δt

α = 8.16rad/s / 4.3s = 1.897 rad/s²,

so tangential a

a= α*r = 1.897rad/s² * 15.24cm

a=28.9cm/s² or = 0.289 m/s²

(b)

velocity

v = ω*r = 8.16rad/s*15.24cm

v= 124.35cm/s = 1.243 m/s

(c) tangential acceleration, like the angular acceleration, is constant while the disk is "spinning up."

a = 0.289 m/s²