A) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diamete

Question

A) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 79.0 rev/min in 3.10 s?

B) When the disk is at its final speed, what is the tangential velocity of the bug?

C) One second after the bug starts from rest, what is its tangential acceleration?

D) One second after the bug starts from rest, what is its centripetal acceleration?

E) One second after the bug starts from rest, what is its total acceleration? (In m/s2 and in degrees from the radially inward direction)

Explanation / Answer

= 79 rev/min * 1min/60s * 2rad/rev = 8.27 rad/s
r = 13in * 0.0254m/in = 0.33 m
= / t = 8.27rad/s / 3.1s = 2.67 rad/s²

(a) at = r = 2.67rad/s² * 0.33m = 0.8811 m/s²
(b) v = r = 8.27rad/s * 0.33m = 2.73 m/s
(c) during spinup, at = 0.8811 m/s²
(d) ac = ²r = (t)²r = (2.67rad/s² * 1s)² * 0.33m = 2.353 m/s²
(e) a = (ac² + at²) = 2.513 m/s²
= arctan(0.8811/2.353) = 20.53º

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