(a) What is the magnitude of the tangential acceleration of a bug on the rim of

Question

(a) What is the magnitude of the tangential acceleration of a bug on the rim of a 12.5-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 77.0 rev/min in 3.60 s? m/s (b) When the disk is at its final speed, what is the magnitude of the tangential velocity of the bug? (c) One second after the bug starts from rest, what is the magnitude of its tangential acceleration? m/s (d) One second after the bug starts from rest, what is the magnitude of its centripetal acceleration? m/s2 (e) One second after the bug starts from rest, what is its total acceleration? (Take the positive direction to be in the direction of motion.) magnitude direction m/s2 ° from the radially inward direction Need Help?Read It

Explanation / Answer

angular accelaration is alpha = (wf-wi)/t

initial angular speed is wi = 0 rad/sec

final angular speed is wf = 77 rev/min = (77*2*3.142)/60= 8.06 rad/s

t = 3.6 sec

then alpha = (8.06-0)/3.6 = 2.24 rad/s^2

tangential accelaration is a_tan = r*alpha

radius r = D/2 = (12.5*0.0254)/2 = 0.158 m

then a_tan = r*alpha = 0.158*2.24 = 0.354 m/s^2

b) v = r*wf = 0.158*8.06 = 1.27 m/sec

c) after one second

final velocity is w = alpha*t = 2.24*1 = 2.24 rad/s

alpha = (2.24-0)/1 = 2.24 rad/s^2

then a_tan = r*alpha = 0.158*2.24 = 0.354 m/s^2

d) ac = r*wf^2 = 0.158*2.24^2 = 0.793 m/s^2

e) a_total = sqrt(ac^2+a_tan^2) = sqrt(0.793^2+0.354^2) = 0.868 m/s^2

direction is theta = tan^(-1)(a_tan/ac) = tan^(-1)(0.354/0.793) = 24 deg from the radially inward direction

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