Why was an axis through the center of mass exculpated in using Eq. 17-29 to determine I? Does this equation apply to such an axis? How can you determine I for such an axis using physical pendulum methods?

Question

Jitender Pal · Answer

Momentum of inertia (I) of a physical pendulum is defined as,
I = T²Mgd/4π²
Here d is the distance of center of mass rom point of suspension, T is the time period of oscillation, M is mass of oscillating object and g is the acceleration due to gravity.
If we include the axis passing through center of mass, then the distance of center of mass rom point of suspension will be zero.
So, d = 0
Substitute the value of d in the equation I = T²Mgd/4π², the momentum of inertial of the pendulum will be zero (I = 0). But for a given mass the momentum of inertia will never be zero. That is why an axis through the center of mass was excluded in using equation I = T²Mgd/4π² to determine I and if the axis will pass through center of mass the above equation (I = T²Mgd/4π²) will not be used
Therefore in this case you can use integration method of continuous system to determine I, which is
I = ∫ dmr²
Here, the body has a small portion which has mass dm and r is the radius.

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Why was an axis through the center of mass exculpated in using Eq. 17-29 to determine I? Does this equation apply to such an axis? How can you determine I for such an axis using physical pendulum methods?

Why was an axis through the center of mass exculpated in using Eq. 17-29 to determine I? Does this equation apply to such an axis? How can you determine I for such an axis using physical pendulum methods?

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Why was an axis through the center of mass exculpated in using Eq. 17-29 to determine I? Does this equation apply to such an axis? How can you determine I for such an axis using physical pendulum methods?

Why was an axis through the center of mass exculpated in using Eq. 17-29 to determine I? Does this equation apply to such an axis? How can you determine I for such an axis using physical pendulum methods?

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