(a) What is the tangential acceleration of a bug on the rim of a 8.0 in. diamete

Question

(a) What is the tangential acceleration of a bug on the rim of a 8.0 in. diameter disk if the disk moves from rest to an angular speed of 80 rev/min in 5.0 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
What is its centripetal acceleration?
______m/s2
What is its total acceleration?
______m/s2
______° (relative to the tangential acceleration)

Please give answers to the blanks provided!

Explanation / Answer

angular velocity attained, ? = 70 rev/min in 4.1 sec ( 4.1/60 min= 0.06833 min ) by newton's equation of motion for angular motion ?final = ?initial + at ?initial= 0 this implies , 70=a*(4.1/60) a = 1024.39 rev/min2 converting into radians/ sec2 a = 1024.39*2p/60*60 =1.78 rad/sec2 tangential acceleration = angular acc.* r = 1.78 * (6/2) =5.36 m/s2 tangential velocity = r* ?(in radians) = 3* 70*2p/60 = 21.99 m/s total accleration = tangential + centripetal ? at one second = a *1 = 1.78 rad/sec centripetal acceleration after 1 sec = r*?( (at one sec)2 = 3*(1.78)2 =9.5052 m/s2 since this acceleration acts along raidial direction and tangential acc acts along tangent there fore angle between at and a centripetal is 90 degrees total acceleration hence = ( (5.36)2 + (9.5052)2 )1/2 = 10.91 m/s2 please rate good

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