(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 79.0 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?

Explanation / Answer

(a)
= (2**79.0)/60 = 8.27 rad/s
f = i + t
8.27 = 0 + *4.0
= 2.06 rad/s^2
Tangential acceleration of a bug, = 2.06 rad/s^2

(b)

r = 13 inches/ 2 = 0.3302/2 = 0.165 m
v = r*f
v = 0.165 * 8.27 m/s
v = 1.36 m/s
Tangential velocity of the bug, v = 1.36 m/s

(c)
at = r*
at = 0.165 * 2.06 m/s^2
at = 0.34 m/s^2
Tangential acceleration, at = 0.34 m/s^2

(d)
vt = vo + at*1
vt = 0 + 0.34*1
vt = 0.34 m/s

ac = vt^2/r
ac = 0.34^2/0.165
ac = 0.7 m/s^2
Centripetal acceleration, ac = 0.7 m/s^2

(e)
Total acceleration, a = sqrt(at^2 + ac^2)
a = sqrt(0.7^2 + 0.34^2)
a = 0.78 m/s^2

Angle = tan^-1(0.34/0.7)
Angle = 25.9o

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