(a) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diamet
ID: 1902302 • Letter: #
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 75.0 rev/min in 4.50 s? m/s/s(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/s/s
(d) One second after the bug starts from rest, what is its centripetal acceleration?m/s/s
(e) One second after the bug starts from rest, what is its total acceleration?m/s/s
Explanation / Answer
(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 75.0 rev/min in 4.50 s?
w = 2f = 2 * 3.1416 * 75/60 = 7.854 rad/s
R = 13 * 0.0254/2 = 0.1651 m
= w/t = 7.854/4.50 = 1.7453 rad/s2
a = R = 0.1651* 1.7453 = 0.28815 m/s2
(b) When the disk is at its final speed, what is the tangential velocity of the bug?
v = R w = 0.1651 * 7.854 = 1.30 m/s
(c) One second after the bug starts from rest, what is its tangential acceleration?
= w/t = 7.854/4.50 = 1.7453 rad/s2
a = R = 0.1651 * 1.7453 = 0.28815 m/s2
(d) One second after the bug starts from rest, what is its centripetal acceleration?
w = t = 1.7453 rad/s
a = R w^2 = 0.1651 * 1.7453*1.7453 = 0.50291 m/s2
(e) One second after the bug starts from rest, what is its total acceleration?
a = (0.28815*0.28815+0.50291*0.50291) = 0.580 m/s2
Arctan(0.576/1.0058) = 29.8° from the radially inward direction
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