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A solid body rotates about a stationary axis according to the law theta=at-bt^3,where a=6 rad/s and b=2 rad/s^3.Find the mean values of the angular velocity and acceleration over the interval between t=0 and the time,when the body comes to rest.

A solid body rotates about a stationary axis according to the law theta=at-bt^3,where a=6 rad/s and b=2 rad/s^3.Find the mean values of the angular velocity and acceleration over the interval between t=0 and the time,when the body comes to rest.

Grade:12

1 Answers

vikas askiitian expert
509 Points
10 years ago

@ = at - bt3               ...............1

d@/dt = W = a-3bt2             ...............2

dW/dt = alfa = -6bt           ............3

body comes to rest when angular velocity  becomes 0

a - 3bt2 = 0

    t = (a/3b)1/2 

mean value of angular velocity = (@t - @0)/t

   @0=0

  @t= a(a/3b)1/2 - b(a/3b)3/2

      = a(a/3b)1/2[1-1/3]

      =(2a/3)(a/3b)1/2

mean value of angular velocity = 2a/3(a/3b)1/2/t = 2a/3 = 4rad/sec

 

now mean value of angular accleration = change in angular velocity/t = (Wt-W0)/t

at t=0  W=a , at T=t  Wt = 0

mean value = a/t = (3ab)1/2 

                 =(36)1/2=6rad/sec2

 

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