Question

A uniform circular disk whose radius R is 12.6cm is suspended as a physical pendulum from a point on its rim. (a) What is its period? (b) At what radial distance r < R is there a pivot point that gives the same period?

Answer 1

The period of a circular disk suspended as a physical pendulum is 2*pi*sqrt(r/g), where r is the radius of the disk and g is the acceleration due to gravity. If we want to find a pivot point that gives the same period, the radial distance r must be equal to the radius of the disk R.

Step-by-step explanation:

To find the period of a circular disk suspended as a physical pendulum, we can use the formula:

T = 2*pi*sqrt(I/mgd)

Where T is the period, I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and d is the distance from the center of mass to the pivot point.

In this case, the moment of inertia of a disk about its axis of rotation is given by:

I = (1/2)*m*r^2

Where r is the radius of the disk.

Plugging in the given values, we get:

T = 2*pi*sqrt((1/2)*m*r^2/mgd) = 2*pi*sqrt(r/g)

(a) The period of the disk is 2*pi*sqrt(r/g).

(b) To find the radial distance r < R that gives the same period, we set two different radii equal to each other and solve for r:

2*pi*sqrt(r/g) = 2*pi*sqrt(R/g)

sqrt(r/g) = sqrt(R/g)

r = R

So, the radial distance r < R that gives the same period is equal to the radius of the disk R.