Question

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 75.0 rev/min in 3.50 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

we need to find angular acceleration

using rotational kinematics

w = w_o + t

= w / t

w = 75 rev/min ( given) = 7.854 rad/sec

so

= 7.854 / 3.50

= 2.243 rad/s²

so,

tangential acceleration

a = r

r = 12.5 in = 0.3175 m

so,

a = 0.7124 m/s²

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(b) magnitude of the tangential velocity of the bug

v = rw

v = 0.3175 * 7.854

v = 2.49 m/s

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(c) tangential acceleration is constant and will be same

so,

a = 0.7124 m/s²

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(d) v_f = 0.7124 * 1 = 0.7124 m/s

so,

a_c = 0.7124² / 0.3175

a_c = 1.598 m/s²

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(e) total acceleration

a = sqrt ( 1.598² + 0.7124²)

a = 1.75 m/s²

and

direction = arctan ( 0.7124 / 1.598)

direction = 24.02 degree