(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 10.5-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 75.0 rev/min in 4.80 s?

(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) Given

Angular Speed, Omega = 75 rev/min = 75*2*pi/60 = 7.849 rad/sec,

Radius of circular path = 5.25 in/39.37 = 0.1365 m

Angular Acceleration, alpha = (omega - omega0)/t = 7.849/4.8 = 1.635 rad/s2

Tangential acceleration = r*alpha = 0.135*1.635 = 0.2232 m/s2

b)

Tangential velocity, v = r*omega = 0.1365*7.849 = 1.0713 m/s

c)

Since both r and alpha are constant the Tangential Acceleration also remains constant ,at t= 1.0 sec, a = 0.2232 m/s2

d)

at t= 1 sec, tangential velocity of bug is 0.2232(1) = 0.2232 m/s

Centripetal acceleration is v^/r = (0.2232)^2/0.1365 = 0.3649 m/s2

e)

Total acceleration = sqrt ( 0.3649^2+0.2232^2) = 0.4277 m/s2

Theta = tan-1 (0.2232/0.3649) = 31.45degrees

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