(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 4.60 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 _______

Explanation / Answer

a)angular speed=76*2pi/60=7.95 rad/sec

angular accleration=7.95/4.6=1.73 rad/s^2

tangential accleration=Ra=1.73*12*2.54/200=0.26 m/s^2 (ans)

b)tangential speed=rw=7.95*12*2.54/200=1.21 m/s (ans)

c)after one sec,tangential accleration is same=0.26m/s^2 (ans)

d)for centrepetal accleration,Wf=1.73 rad/sec

centrepetal acc=Wf^2R=1.73*1.73*12*2.54/200=0.456 m/s^2 (ans)

e)total accleration=sqrt(At^2+Ac^2) =0.53m/s^2 (ans)

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