(a) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diamet
ID: 2182183 • 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/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
d = 13*2.54 = 33.02 cm
r = d/2 = 16.51 cm = 0.1651 m
w = 75*2*pi/60 = 7.85 rad/s
a) alfa = (w-wo)/t
= (7.85-0)/4.3
= 1.826 rad/s^2
a_tan = r*alfa
= 0.1651*1.826
= 0.3 m/s^2
b) v_tan = r*w
= 0.1651*7.85
= 1.296 m/s
c) a_tan = 0.3 m/s^2
d) a_rad = r*w^2
= 0.1651*(alfa*t)^2
= 0.1651*(1.826*1)^2
= 0.55 m/s^2
e) a = sqrt(a_tan^2 + a_rad^2)
= sqrt(0.3^2 + 0.55^2)
= 0.626 m/s^2
theta = tan^-1(a_tan/a_rad)
= tan^-1(0.3/0.55)
= 28.6 degrees
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