Question

If a disc is made to oscillate from a point R/4 away from centre and the axis of oscillation is perpendicular to the plane of disc then the length of equivalent simple pendulum will be

Answer 1

Answer:

A disc of radius R and mass M is pivoted at the rim and set for small oscillations about an axis perpendicular to plane of disc. If simple pendulum have

Answer 2

Answer:

9/4R

Explanation:

the moment of inertia of the disc perpendicular to the plane passing through the center (Id)=MR^2/2

where,

M=mass of the disc

R=radius of the disc

as given in question the disc is made to oscillate at  R/4 away from the center

so we need to find the new moment of inertia of the disc at R/4 from the center by using the parallel axis theorem

Id'=MR^2/2+M(R/4)^2=9/16MR^2

now the time period of the disc(Td)=2π$\sqrt{I/Mgd}$         (1)

where I=moment of inertia of the disc

M=mass of the disc

d=distance of the axis of oscillation from the center of the disc

g=acceleration due to gravity

the time period of a simple pendulum(Tp)=2π$\sqrt{L/g}$             (2)

L=length of the pendulum

g=acceleration due to gravity

by equating equations 1 and 2 we get:

2π$\sqrt{L/g}$=2π$\sqrt{Id'/Mgd}$

by substituting the values we get

L=9/4R