(a) What is the tangential acceleration of a bug on the rim of a 11.0-in.-diamet

Question

(a) What is the tangential acceleration of a bug on the rim of a 11.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 76.0 rev/min in 4.80 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?

Your response differs from the correct answer by more than 100%. m/s2

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

m/s and

Explanation / Answer

r=11/2= 5.5inch= 0.14m

w1=0rad/s , w2= 76rev/min= (76*2)rad/60s = 7.95rad/s

t=4.80s

a= (w2-w1)/ t = (7.95-0)/4.80 = 1.66rad/s^2

a) at= a*r = 1.66*0.14 = 0.2324 m/s^2

b) v= w2*r = 7.95*0.14 = 1.11 m/s

c) After 1s, since the disk accelerates uniformly angular acceleration will be same as before

hence a=1.66rad/s^2

at= a*r = 1.66*0.14 = 0.2324 m/s^2

d) after 1s,

ac= v^2/r = (0.2324^2)/0.14 = 0.39 m/s

e) he total acceleration for circular motion is given by the vector sum of the centripetal and tangential accelerations

|a| = sqrt[at^2+ac^2) = sqrt(0.2324^2+0.39^2) = 0.45 m/s^2

Direction of ‘a’ with respect to ac = q = tan^-1(at/ ac) = tan^-1(0.2324/0.39) = 30.8 deg

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.