(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 13.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 80.0 rev/min in 4.80 s? ____ m/s2

(b) When the disk is at its final speed, what is the magnitude of the tangential velocity of the bug? ____m/s (c)

One second after the bug starts from rest, what is the magnitude of its tangential acceleration? _____ m/s2

(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____ m/s2

direction _____ ° from the radially inward direction

Explanation / Answer

a)

Angular Speed, Omega = 80 rev/min = 80*2*pi/60 = 8.373 rad/sec,

Radius of circular path = 6.5 in/39.37 = 0.1651 m

Angular Acceleration, alpha = (omega - omega0)/t = 8.373/4.8 = 1.744 rad/s2

Tangential acceleration = r*alpha = 0.1651*1.744 = 0.2879 m/s2

b)

Tangential velocity, v = r*omega = 0.1651*8.373 = 1.382 m/s

c)

Since both r and alpha are constant the Tangential Acceleration also remains constant ,

at t= 1.0 sec, a = 0.2879 m/s2

d)

at t= 1 sec, tangential velocity of bug is 0.2879(1) = 0.2879m/s

Centripetal acceleration is v^2/r = (0.2879)^2/0.1651 = 0.5020 m/s2

e)

Total acceleration = sqrt ( 0.5020^2+0.2879^2) = 0.578 m/s2

Theta = tan-1 (0.2879/0.5020) = 29.83 degrees

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