(a) What is the tangential acceleration of a bug on the rim of a 12.0-in.-diamet

Question

(a) What is 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 76.0 rev/min in 4.50 s? m/s2 (b) When the disk is at its final speed, what is the tangential velocity of the bug? m/s (c) One second after the bug starts from rest, what is its tangential acceleration? m/s2 (d) One second after the bug starts from rest, what is its centripetal acceleration? m/s2 (e) One second after the bug starts from rest, what is its total acceleration? m/s2

Explanation / Answer

The tangential speed is vt = ?*r (a) vt0 = ?0*r vt1 = ?1*r ?vt = vt1 - vt0 = (?1 - ?0)*r tangential acceleration at = ?vt/?t ?0 = 0, ?1 = 76 RPM = 7.96 rad/s r = 6 in = 0.152 m ?vt = 7.96*0.152 = 1.21 m/s at = 1.21/3 = 0.40 m/s² (b) vt1 = ?1*r = 1.21 m/s (c) if the angular acceleration is uniform, then the tangential acceleration is constant. the centripetal acceleration is ?²*r. after 1 s, the angular velocity is 1/3 the final velocity at 3 s. so ? = 7.96/3 = 2.65 rad/s and the centripetal acceleration is 2.65²*0.152 = 1.07 m/s² The total acceleration is the sq rt of the sum of the squares of centripetal and tangential accelerations: a = v[1.07² + 0.40²] = 1.14 m/s² The angle rel to at is arctan (ac/at) = arctan (1.07/0.4) = 69.5º

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