(a) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diamet

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 80.0 rev/min in 4.00 s? m/s2 (b) When the disk is at its final speed, what is the tangential velocity of the bug? m/s (c) One second after the bug starts from rest, what is its tangential acceleration? m/s2 (d) One second after the bug starts from rest, what is its centripetal acceleration? m/s2 (e) One second after the bug starts from rest, what is its total acceleration? m/s2

Explanation / Answer

similar problem (a) What is the tangential acceleration of a bug on the rim of a 9.0 in. diameter disk if the disk moves from rest to an angular speed of 80 revolutions per minute in 5.0 s? m/s2 (b) When the disk is at its final speed, what is the tangential velocity of the bug? m/s (c) One second after the bug starts from rest, what is its tangential acceleration? m/s2 What is its centripetal acceleration? m/s2 What is its total acceleration? m/s2 ° (relative to the tangential acceleration) 9.0 in = 0.2286 meters 80 revolutions per minute = 80/60*2pi= 8.37758 rad/sec (a) angular acceleration = (8.37758 rad/sec)/5.0 s = 1.675516 rad/s^2 tangential acceleration = 1.675516*0.2286 = 0.383 m/s^2 (b) tangential acceleration = 0 m/s^2, there is no more angular acceleration. (c) tangential acceleration = 0.383 m/s^2; same as (a) tangential speed = tangential acceleration * time = 0.383 m/s^2 * 1 sec = 0.383 m/s a = (v^2)/r = (0.383^2)/0.2286 = 0.642 m/s^2 total acceleration = sqrt(0.383^2 + 0.642^2) = 0.747 m/s^2 direction relative to the tangential acceleration = tan^-1(0.642/0.383) = 59.2 degrees towards the center.

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