(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 3.60 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
° from the radially inward direction

Explanation / Answer

angular speed = = 79rev/min = (79*2)/60 = 8.273 rad/s

a) angular acceleration, = (8.273-0)/(3.60) = 2.298 rad/s2
tangential acceleration, a = r = 6.5*2.298 = 14.937 in/s2 = 0.0254*14.937 = 0.3794 m/s2

b) when the disk is at final speed , = (79*2)/60 = 8.273 rad/sec
tangential velocity, v = r = 6.5*0.0254*8.273 = 1.3659 m/s

c. tangential acceleration, a = r = 6.5*2.298 = 14.937 in/s2 = 0.0254*14.937 = 0.3794 m/s2

d) After 1 sec from rest,
= 0+t => = 2.298*1 = 2.298 rad/sec
v = r = 6.5*0.0254*2.298 = 0.3794 m/s
centripetal acceleration = v2/r = 0.3794^2/6.5*0.0254 = 0.872 m/s2

e)after 1 sec , centripetal acceleration = 0.872 m/s2 and tangential acceleration = 0.3794m/s2
total acceleration = (0.872^2 +0.3794^2) = 0.951 m/s2

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