Question

Q3. For the rotational system subjected to an applied torque Mocosout shown in Figure 3, the rotary inertia of the rigid bar about the hinge O can be calculated by Jo =7ml /48. Given k = 5,000N/m, 1 - 1m, m = 20 kg, Mo = 100 Nm, c = 130 rpm. Assume rotation angle is very small, (i) Draw the free body diagram; (ii) Use Newton's 2nd law to derive the equation of motion of the system; and (iii) Find the natural frequency and natural period. Uniform rigid bar. mass Mocos ou c00000 Figure 3

Answer 1

(i)

Here, in the absence of the applied torque, the gravitational force still will be acting on the system. As a result there will be some initial elongation in the right spring and respective compression in the left spring to counterbalance the effect of this gravitational force.

When the torque is applied, the springs are therefore already deformed due to gravitational force. Considering this initial deformed position as the initial position of the system, the effect of gravity can be neglected in our analysis.

kxı sin kxi cos e - - - - - M, cos wt - KI2 - -- - kxi | HS X1 - - - - 1/4 31/4 Qo 01 22

(ii)

The torque applied by the right spring about the hinge 'O' will be given as:

(kz1 cos) < (31/4)

Since the rotation angle $\theta$ is very small,

cos 1

and,

21 = 31/4 x 0

Thus, the torque applied by the right spring can be given as:

k (31/4) a < (31/4) = k (31/4)

Similarly, the torque applied by the left spring can be given as:

k (1/4) x (1/4)= k (1/4)

From Newton's 2nd law of motion,

Tort = J.

M.coswt - k (31/4) - k (1/4) = J.ä

= JQ+k (912/16) +k (1/16) 0 = M, coswt

→ Jö+k (1012/16) 6 = M, coswt

= (and') ö + (!!) – Mo casu

$\Rightarrow \ddot{\theta}+\left ( \frac{60kl^2}{14ml^2} \right )\theta=\left (\frac{48M_o}{7ml^2} \right )\cos\omega t$

The above equation is the equation of motion of the given system.

(iii)

From the above equation of motion, the natural frequency of vibration of the system can be calculated as:

60612 n = V 14ml

Substituting the respective values;

n = V 60 x 5000 x 12 14 x 20 x 12

wn = 32.732 rad) sec.

which is the natural frequency of the system.

Also,

น” ..

27 :.T = —

Thus, natural time period of the system can be calculated as:

2π 32,732 που = 0.192 sec.