An electric turntable 0.800 m in diameter is rotating about a fixed axis with an

Question

An electric turntable 0.800 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.290 rev/s . The angular acceleration is 0.885 rev/s2 .

a.) Compute the angular velocity after a time of 0.206 s .

b.) Through how many revolutions has the blade turned in this time interval?

c.) What is the tangential speed of a point on the tip of the blade at time t = 0.206 s ?

d.) What is the magnitude of the resultant acceleration of a point on the tip of the blade at time t = 0.206 s ?

Explanation / Answer

Given

   diameter d = 0.8m , radius r = 0.4 m

initial angular velocity u = 0.290 rev/s

angular acceleratin a = 0.885 rev/s2

time t = 0.206 s

a)angular velocity after t = 0.206 sis

   v = u+at

   v = 0.290+0.885*0.206rev/s = 0.47231 rev/s = 0.47231/(2pi) rad/s = 0.07517 rad/s

b)Through how many revolutions has the blade turned in this time interval?

   using equations of motion s

   v^2 - u^2 = 2a*X
  
   0.47231^2 - 0.290^2 = 2*0.885*x

   X = 0.07851793 rev

  
c)   Tangential speed is V = r*w = 0.4*0.47231/2pi m/s = 0.0300682 rad/s

d) resultant acceleration is

   a = r*alpha+ w^2*r

   a tan = 0.4*0.885 = 0.354 m/s2

   a rad = W^2*r =0.07517^2*0.4 m/s2 = 0.002260 m/s2

now magnitude isa = sqrt((0.354)^2+(0.002260)^2) m/s2 = 0.3540 m/s2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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