The period T of a physical pendulum is measured about a certain pivot point A. T

Question

The period T of a physical pendulum is measured about a certain pivot point A. The measurement is repeated, varying the locations of the pivot point but keeping the rotations of the pendulum always in the same plane. The moment of inertia of the object about its center of mass is ICM, the mass is M, and the distance from the pivot point to the CM is d. Find the period as a function of ICM, M, d, and the acceleration due to gravity (g). Note that as the pivot point is varied, pivot points that are the same distance from the CM give the same period. Furthermore, show that, in general for a given value of T, there are two distances d1 and d2 that give the same period of oscillation. Show that the moment of inertia of the object about its center of mass and the period can be expressed as [DIFFICULT] Show that the smallest period for rotations occurs when the distance of the CM from the pivot is d =

Explanation / Answer

a)T=2*pi*sqrt(I/(M*g*d)

in the eqn above,I is inertia of mass about pivot point which can be rewritten as I=Icm+M*d^2

parallel axis theorem

d is the distance pivot point and centre of mass

B)if T is constant, then for d,

(T/2*pi)^2=(Icm+M*d^2)/M*g*d

let(T/2*pi)^2=k ,a constant as T is constant

now expressing d interms of k,Icm,M,g we get

M*d^2+M*g*d*k^2+Icm=0

this is a quadriatic equation in d so it has two roots say d1 and d2

So there are two distances d1 and

d2 that give the same period of oscillation.

C)From the above quadriatic equation d1 and d2 are distances and roots of the equation

we can clearly see that product of roots of above equation is Icm/M

So d1*d2=Icm/M,so

M*d1*d2=Icm

D)For Smallest period,differentiate T=2*pi*sqrt(I/(M*g*d) with d and equate to zero

differentiatiing we get d=sqrt(Icm/M).

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