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 center of mass.For a circular disk, $I _ { \mathrm { cm } } = \frac { 1 } { 2 } m R ^ { 2 }$ .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 center of mass.For a circular disk, $I _ { \mathrm { cm } } = \frac { 1 } { 2 } m R ^ { 2 }$ .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

As given, the angular frequency for small oscillations is approximately equal to $\sqrt{\frac{g}{R}}$. Substituting the value of g and R we get, $\sqrt{\frac{9.81}{1.0}} = 3.13$ rad/s. However, this is the angular frequency for oscillations about the center of mass. For oscillations about the pivot, we need to account for the displacement of the center of mass from the pivot. The effective length of the pendulum is given by $l = d + R = 1.25$ m. Using this in the above formula, we get $\sqrt{\frac{9.81}{1.25}} = 2.21$ rad/s. This is closest to option B, 2.1 rad/s.

In the Figure Below,a Disk (Radius R = 1 Icm=12mR2I _ { \mathrm { cm } } = \frac { 1 } { 2 } m R ^ { 2 }Icm​=21​mR2

In the Figure Below,a Disk (Radius R = 1 $I _ { \mathrm { cm } } = \frac { 1 } { 2 } m R ^ { 2 }$