Help with Dynamics Rotor's Angular Velocity??

Question

The problem states "The Heclicopter's rotor starts from rest. The moment exerted on it (in N-m) is given as a function of the angle through which it has turned in radians by M = 6500 - 20 x theta. The rotor's moment of inertia is I = 540 kg-m^2. Determine the rotor's angular velocity (in rpm) when it has turned through 10 revolutions." The answer in the back of the book is 353 rpm (37.0 rad/s) but i'm not sure how to set this problem up. Any help would be great!

Answer 1

Edited Answer:

The angular acceleration α of the rotor is given by T = I*α.  So α = T/I   T is given as a function of θ:  T(θ) = T0 - k*θ.  ÃƒÂŽÃ‚± is therefore not a constant.

α = d²θ/dt²

α = T/I

T/I =  d²θ/dt²

T = T0 - k*θ

(T0 - k*θ)/I = d²θ/dt²

d²θ/dt² + (k/I)*θ  - T0/I = 0

The general solution to this differential eq is

θ(t) = C1*sin√[k/I]*t + C2*cos√[k/I]*t

At t = 0, dθ/dt = 0 so C1 = 0

θ(t) = C2*cos√[k/I]*t

The particular solution is a constant K, so the complete solution would be θ(t) = C2*cosh√[k/I]*t + K.  Put that into the original differential equation to get:

-C2*(k/I)*cosh√[k/I]*t + C2*(k/I)*cosh√[k/I]*t +(k/I)* K - T0/I   = 0,

the  first two terms cancel, so

(k/I)* K = T0/I;   K = T0/k

θ(t) = C2*cos√[k/I]*t + T0/k

If θ at t = 0 is taken to be 0, then C2 = -T0/k

θ(t) =  T0/k*[1 - cos√[k/I]*t]

Find t for θ = 20π, then find ω= dθ/dt at that value of t.

Answer 2

Let angular velocity  =  w

So angular acceleration   =   dw/dt

Let angle rotated   =   A

So    w   =   dA/dt

Use    dw/dt   =   dw/dA   x   dA/dt

So      dw/dt   =   dw/dA   x   w

i.e.    dw/dt   =   w*dw/dA

If  "C" is the couple acting, "I" is the moment of inertia, and "dw/dt" the angular acceleration, then:-

                         C   =   I * dw/dt

Substituting values from question and above:-

              6500  -  20 * A    =    540  *  dw/dt

    So          6500 -  20 * A    =    540  *  w  *  dw/dA

Tnis gives:   (6500  -  20*A)dA     =     540*w*dw

Integrating     6500*A   -   20*A^2/2    =    540 *w^2/2   +   C

  At start,   A   =   0  when   w   =   0,  so C  =  0

This gives

                  6500*A   -   10*A^2    =    270*w^2

This equation gives the angular speed "w" after  turning through angle "A" radians.

       Here,   A  =  10  turns   =   20*pi   radians

      So     6500 * 20 * pi  -  10*(20*pi)^2   =    270 * w^2

     This gives   w   =   36.96 rads/sec

                    Say    37 rads/sec.

  (the trick appears to be the line    dw/dt   =   dw/dA  *  dA/dt  )