(a) What is the tangential acceleration of a bug on the rim of a 12.0-in.-diamet

Question

(a) What is the tangential acceleration of a bug on the rim of a 12.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 76.0 rev/min in 4.20 s?
m/s^2

(b) When the disk is at its final speed, what is the tangential velocity of the bug?
m/s^2

(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) a= alpha r

= ((76*2pi/60)/4.2)* 0.3048/2

= 0.289 m/s^2

B) v = at = 0.289*4.2= 1.214 m/s

C) a= 0.289 m/s^2

D) ac = v^2 /r

= (0.289)^2 /0.1524

= 0.548 m/s^2

E) Total acceleration = sqrt(0.548^2 +0.289^2)

=0.62 m/s^2

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