(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 4.90 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) wf -wi = alpha x t

(79 x 2pi/60 rad/s) - 0 = alpha x 4.90

alpha = 1.69 rad/s2

a_t = alpha x r

r = 13/2 inch = 6.5 x 0.0254 = 0.1651 m

a_t = 1.69 x 0.1651 = 0.279 m/s^2

b) v = w *r

   = (79x 2pi / 60 rad/s) (0.1651 m) = 1.37 m/s

c) acc. is same throught.

a = 0.279 m/s2

d) aftre 1 sec

V = at = 0.279 x 1 = 0.279 m/s

a_c = v^2 / r = 0.279^2 / 0.1651 = 0.471 m/s2

e) a_total = sqrt(a_c^2 + a_t^2) = 0.548 m/s2

direction = tan-1(a_t /a_c) = 30.64 degrees

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.