What is the tangential acceleration of a bug on the rim of a 13.0-in.-diameter d

Question

What is 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 79.0 rev/min in 4.90 s?

Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2

(b) When the disk is at its final speed, what is the tangential velocity of the bug?

(c) One second after the bug starts from rest, what is its tangential acceleration?
Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2

(d) One second after the bug starts from rest, what is its centripetal acceleration?

Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2

(e) One second after the bug starts from rest, what is its total acceleration?

Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2   
Your response differs from the correct answer by more than 10%. Double check your calculations.

Explanation / Answer

angular speed W = 79 rpm

and 1 rpm = 2pi/60 rev/sec

so

W = 79* 2pi/60

W = 8.27 rad/sec

apply the kinematic equation Wf = Wo + alpha T

where alpha is angular accleration

t is time

so

alpha = Wf-Wo/t

alpha = (8.27-0)/(4.9)

alpha = 1.678 rad/s^2

now use a = r alpha

1 in = 0.0254 mm

here diamter D = 13 in = 13/2 = 6.5 in = 6.5 *0.0254 = 0.1651 m

a = 0.1651 * 1.678

a = 0.277 m/s^2

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b. use V = r W

V = 0.1651 * 8.27

V = 1.36 m/s

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a = rW

a   = 0.277 m/s^2

----------------------

d. alpha = ang accleration   = 1.678 rad/s^2

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