In the system shown below, the rigid, massless bar of length \'l\' rotates about
ID: 1940307 • Letter: I
Question
In the system shown below, the rigid, massless bar of length 'l' rotates about "O". The motion of the mass'm' is excited with an initial displacement of x(0) = 0 and an initial velocity of x(0) = 0.1 m/s. Determine the equations of motion of the system Compute the natural frequency and damped frequency at which the system will vibrate. Which of the following graphs is the correct response of x(t): (i), (ii) or (in) Keeping the damping coefficient 'c' and the spring stiffness 'k' at the same values as given above, what would the mass 'm' have to be changed to in order for the system to be critically damped.Explanation / Answer
A.
Since the link is a mass-less body, it will not affect the equations.
Note that when the body (m) moves a distance (x), the spring will be stretched by distance (x), and the damper by (x/2). To get the equation of motion, we will get moments around point (O):
(m*a)*l/2 + (c*v)*l/4 + (k*x)*l/2 = Jo*
For massless body (Jo = zero), dividing by (l/2):
m*xoo + c*xo/2 + k*x = 0
Where xoo is the second derivative of distance x, or acceleration.
And xo is the first derivative of distance x, or velocity.
B.
Get it in the (S) domain:
m*xoo + c*xo/2 + k*x = 0
m*[s2x(s) - sx(0) - xo(0)] + c*[sx(s) - x(0)] + k*[x(s)] = 0
For x(0) = 0:
m*[s2x(s) - xo(0)] + c*[sx(s)] + k*[x(s)] = 0
Take x(s) as a common factor:
x(s)*[m*s2 + c*s + k] = m*xo(0)
x(s) = [m*xo(0)]/[m*s2 + c*s + k]
Divide by m:
x(s) = [xo(0)]/[s2 + c*s/m + k/m]
Where:
(k/m) is the natural frequncey wn = (32/2) = 4 Hz
And (c/m) = 2**wn , where wd = wn*(1-2)
so: = 6.4/(2*2*4) = 0.4
wd = 4*(1- 0.42) = 3.67 Hz
C.
I can't see any of the graphs clearly, but no.(iii) is the most logical one.
D.
For critical damping: = 1
Since: And (c/m) = 2**wn , wn = (k/m)
(c/m) = 2**(k/m)
By squaring the equation:
(c/m)2 = 4*2*(k/m)
For c = 6.4, k = 32, = 1:
m = c2/(4*k) = 6.42/(4*32) = 0.32 kg