(a) What is the tangential acceleration of a bug on the rim of a 12.0-in.-diamet
ID: 2013446 • 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 77.0 rev/min in 4.00 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
_______° from the radially inward direction
Explanation / Answer
In order to find the tangential acceleration, you first need to edit the angular speed so that it is in the form rad/s. you do that by multiplying it by (2pi rad/1 rev) because 1 revolution is equal to 2pi, or around the circle. Then multiply that by (1 min/60 sec) to get 8.0634 rad/sec Now using w=w0 + (alpha)t you get 8.0634rad/s=0rad/s + (alpha=angular acceleration)(4s) angular acceleration=2.01585rad/s2 and tangential acceleration is equal to angular acceleration times the radius which radius is half of the 12 inches. 6inches times (1 m/(3*12inches))=.167 m so tangential acceleration =.33664695 m/s2 the tangential velocity is equal to angular velocity times the radius. so (8.0634rad/s)(.167m)=1.3466m/s I cannot figure out how to get the next part, but i do know that you need to find the angular velocity after that 1 second. With that you can find the angular acceleration, which leads to the tangential acceleration. the centripital acceleration is found by using the angular velocity squared times the radius. and the total acceleration is the square root of angular acceleration squared plus the tangential acceleration squared. Good Luck. and i believe the angle is inverse tangent of one over the other, i'm not sure which though.
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