(a) What is the tangential acceleration of a bug on the rim of a 12.0-in.-diamet
ID: 1967728 • Letter: #
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 75.0 rev/min in 3.20 s?
1 m/s2
(b) When the disk is at its final speed, what is the tangential velocity of the bug?
2 m/s
(c) One second after the bug starts from rest, what is its tangential acceleration?
3 m/s2
(d) One second after the bug starts from rest, what is its centripetal acceleration?
4 m/s2
(e) One second after the bug starts from rest, what is its total acceleration?
5 m/s2
6° from the radially inward direction
5 m/s2
6° from the radially inward direction
Explanation / Answer
a) so the formula you use is = 0 + t (for angular velocity using angular acceleration)
= 75 rev/min times 2 rad/rev = 471.2 rad/min times 1 min/60 s = 7.85 rad/s
= 0 + t
0 is 0 because it starts from rest.
7.85 = 0 + (3.20)
= 2.45 rad/s2
but it asks for tangential acceleration so multiply by the radius
radius = 12 in/2 times 2.54 cm/1 in times 1 m/100 cm = .1524 m
a = r
a = (.1524)(2.45) = .374 m/s2
b) we already figured out was the final angular velocity which is 7.85 rad/s
multiply by the radius again to get tangential velocity
v = r
v = (.1524)(7.85) = 1.20 m/s
c) since it says uniform acceleration in part a, the acceleration here is the same as part a which is .374 m/s2
d) use = 0 + t to get the tangential velocity after 1 second and use ac = 2r to get the centripetal acceleration.
= 0 + t
= 2.45 from part a, and t = 1.00 s
= 0 + 2.45(1.00) = 2.45 rad/s
ac = 2r = (2.45)2(.1524) = .915 m/s2
e) total acceleration is the square root of the sum of the squares of the tangential acceleration and centripetal acceleration
atotal = (ac2 + at2)
ac = .915 m/s2
at = .374 m/s2
((.915)2 + (.374)2) = .988 m/s2
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