(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 moves from rest to an angular speed of 76 rev/min in 4.0 s?
(b) When the disk is at its final speed, what is the tangential velocity of the bug?
(c) One second after the bug starts from rest, what is its tangential acceleration?
d) What is its centripetal acceleration?
e)What is its total acceleration?

I figured out the answers up until part c, but I had a question about that part.. I was wondering why the the tangential acceleration one second after the bug starts from rest is the same as the value for tangential acceleration in part A. Also, I am completely stumped on centripetal acceleration, I tried the equation ac= v^2/r and I couldn't get the right answer. if you could help I would really appreciate it!!

Explanation / Answer

a.

= /t = (2 76 rad/min)/(4.0 s) = 1.99 rad/s2

a = r = (1.99 rad/s2)(12.0/2 in) = 0.303 m/s2

b.

v = r = (2 76 rad/min)(12.0/2 in) = 1.21 m/s

c.

Since the disk is undergoing constant acceleration, both the angular acceleration and linear acceleration are constant during that period. Therefore, the answer remains 0.303 m/s2.

d.

= t = (1.99 rad/s2)(1 s) = 1.99 rad/s

ac = v2 / r = (r)2 / r = 2r = (1.99 rad/s)2(12.0/2 in) = 0.604 m/s2

e.

((0.303 m/s2)2 + (0.604 m/s2)2) = 0.676 m/s2

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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