a) What is the tangential acceleration of a bug on the rim of a 12.0 in. diamete

Question

a) What is the tangential acceleration of a bug on the rim of a 12.0 in. diameter disk if the disk accelerated uniformly from rest to an angular speed of 80.0 rev/min in 4.90s?

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) one second after the bug starts from rest, what is its centripetal acceleration?

e) one second after the bug starts from rest, what is its total accerleration?

Explanation / Answer

(a) given n = 80 rpm = 80/60 = 1.33 rev/sec
Thus by ? = 2?n = 2 x 3.14 x 1.33 = 8.37 rad/sec
Thus by ?(final) = ?(initial) + ?t
=>8.37 = 0 + 4.9?
=>? = 1.708 rad/sec^2
Thus by a = r? = (12/2) x 2.54 x 10^-2 x 1.708 {as 1in = 2.54 cm & 1 cm = 10^-2m}
=>a = 0.26 m/s^2
(b) By v = r? = (12/2) x 2.54 x 10^-2 x 8.37 = 1.275 m/s
(c) By a = r? = 0.26 m/s^2
(d) angular acceleration(?) = 1.708 rad/sec^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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