Question

One simple pendulum and the physical pendulums (disk and rod) are suspended on the crossbar, as shown in figure. (a) Calculate the natural linear frequency of the simple pendulum, if the length of the simple pendulum is =1.6 m (b) Calculate the natural angular frequency of the disk. The radius L 5 of the disk is R=0.5 m; moment of inertia about an axis through the 0.3 R center of mass is ICM =mR2 (c) Calculate the natural period of oscillations of the rod. The length of the rod is L-1.5 m; moment of inertia about an axis through the center of mass is IcM 12

Answer 1

a)

For simple pendulum, natural linear frequency is

- 1 3 1 - 2π/ί 1 9.81 , 2πV 1.6 1 - = 0.394 Hz

b)

Time period of physical pendulum is

T = 271 mgd

Distance from pivot to center of mass of disk

d=R

Moment of inertia of disk about the pivot point is

I = ICM + md= -mR? + mR = -mR

Angular frequency of disk  $\omega=\frac{2\pi}{T}=\sqrt{\frac{mgd}{I}}=\sqrt{\frac{mgR}{\frac{3}{2}mR^2}}=\sqrt{\frac{2g}{3R}}=\sqrt{\frac{2*9.81}{3*0.5}}=3.62\,rad/s$

c)

Distance from pivot to center of mass of rod is  $d=\frac{L}{2}$

Moment of inertia of rod  about the pivot point is

1 = lemn + md = ime? + m () = 5m2= I = ICM + md = 3

Time period of rod is

T = 271 m2 mg / 2* 1.5 V 3 * 9.81 = 0.319 s mgd