In the figure below, a disk (radius R = 1.0 m, mass = 2.0 kg) is suspended from a pivot a distance d = 0.25 m above its centre of mass. For a circular disk, . The angular frequency (in rad/s) for small oscillations is approximately:
A)4.2.
B)2.1.
C)1.5.
D)1.0.
E)3.8.

Question

In the figure below, a disk (radius R = 1.0 m, mass = 2.0 kg) is suspended from a pivot a distance d = 0.25 m above its centre of mass. For a circular disk,  . The angular frequency (in rad/s) for small oscillations is approximately: 
A)4.2. 
B)2.1. 
C)1.5. 
D)1.0. 
E)3.8.

Quizplus · Accepted Answer

The angular frequency for small oscillations of an object in simple harmonic motion is given by $\omega=\sqrt{\frac{g}{l}}$, where $g$ is the acceleration due to gravity and $l$ is the effective length of the pendulum (the distance between the pivot point and the centre of mass of the object). In this case, $g=9.8$ m/s$^2$ and $l=d+R=1.25$ m. Substituting these values, we get $\omega=\sqrt{\frac{9.8}{1.25}}\approx2.1$ rad/s. Thus, the best choice is B.

In the Figure Below, a Disk (Radius R = 1