(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 76.0 rev/min in 4.60 s? 0.29 m/s2 (b) When the disk is at its final speed, what is the tangential velocity of the bug? 1.3 m/s (c) One second after the bug starts from rest, what is its tangential acceleration? 0/29 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

76 rev / min = 76/60 * 2 = 3.14 rad/s

Also, 13 in = 0.3302 meters

(a) Angular acceleration: 3.14 rad/sec / 4.6 sec = 0.68 rad/sec2

Tangential Acceleration: 0.68 rad/sec2 * 0.3302 = 0.29 m/s2

(b) At its final speed, tangential acceleration becomes 0 m/s2 and hence, velocity of the bug remains constant. (=1.3 m/s)

(c) 1 sec afterwards, tangential accelerations remains the same, i.e. 0.29 m/s2

(d) Centripetal acceleration = a = v2/r

tangential speed (v) = tangential acceleration * time = 0.29 * 1 sec = 0.29 m/s

Hence, a = 0.0841/0.1651 = 0.50 m/s2

(e) Total acceleration = Vector summation of the tangential acceleration and the centripetal acceleration

= sqrt ( 0.292 + 0.502)

at an angle of 90 - tan-1(0.50/0.29) to the radially inward direction (vector triangle)

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