Question

A thin, uniform disk with mass M and radius R is pivoted about a fixed point a distance x from the center of mass. The disk rotates around the fixed point, with the axis of rotation perpendicular to the plane of the disk. What is the period as a function of x, M, R, and acceleration due to gravity (g), and what is the distance x that gives the smallest period?

Answer 1

The time period foe a disk is given by T = 2*pie* ( I/k)^1/2

here I = moment of interia and k = torsional constant = torque/@ = mgx sin@ / @

k (approx equal to) = mgx                      ------------------1

Let P is the pivoted point as shown in fig. so moment of interia about pivoted point is:

                      -------------------------2

For small value of @ time period will be:

                  ( from eq 1 and 2)

(B) for smallest time period dT/ dx = 0

d( 2*pie( R² + 2x² / 2gx)^1/2 / dx = 0

By using formula d( u.v) = ( v * du - u* dv) / v²

2pie( 2gx* ( 4x / 2* (R² + 2X² ) ^1/2 - (R^2 + 2x^2)^1/2 *2g) / (2gx) ^2 = 0

4*g*x² = (R² + 2x² ) * 2g

R =0    ( radius =0)

hence disk must be placed at the center to get minimum time period