Question

(a) What is the magnitude of the tangential acceleration of a bug on the rim of a 12.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 80.0 rev/min in 3.60 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

Answer 1

Given

diameter of the disk is d = 12.0 in ==> radius r = 6 in = 0.1524 m

initial angular velocity wi = 0 rad/s

final angular speed is wf = 8 rev/min in 3 sec = 8*2pi/60 rad/s = 0.838 rad/s

a)

now the tangential acceleration is a_t = r*alpha = r*(wf-wi/t) = r(wf-0)/t

substituting the values

a_t = 0.1524*0.279 m/s2 = 0.0425196 m/s2

b) tangential velocity is v_t = r*Wf = 0.1524*8*2pi/60 = 0.1277 m/s

c)One second after the bug starts from rest, what is the magnitude of its tangential acceleration

as the disk is accelerating uniformly the tangential acceleration is same as it is at 3 seconds

a_t = 0.0425196 m/s2

d) what is the magnitude of its centripetal acceleration a_cen = ?

We know that the a_cen = v_t^2/r

a_cen = 0.1277^2 /0.1524 m/s2 = 0.1070 m/s2

e) total accelration is a = sqrt(a_t^2+a_cen^2) = sqrt(0.0425196^2+0.1070^2) m/s2 = 0.115 m/s2

the direction is tan theta = a_t/a_cen = 0.0425196/0.1070

theta = 21.672 degrees radially inward direction