A threatening note written on 85x11 inch paper is pinned to

A threatening note written on 8.5x11 inch paper is pinned to the door of a physics professor. The perpetrator left in a hurry so that when the physics professor finds the paper it is still swinging. Assuming the pin is at the very top center of the paper, what is the period of the paper’s motion?

Solution

We know that for an elongated body of Inertia I and mass M, the time period of oscillations is given as

T = 2 sqrt(I / Mga) where a is the distance of centre of mass to the point for T is to be determined.

Also, for the note to be hanging by a pin passing through the top centre of the paper, we can consider the paper as a thin rectangular sheet of length a, breadth b and mass M.

Hence the inertia of the paper about the pin = M(a^2 + b^2)/12 + Ma^2/4 = M(4a^2 + b^2) / 12

That is Inertia of the paper = M(4(121) + 8.5*8.5) / 12*144 [We divide by 144 to change inches to feet]

I = 0.3219 M

Putting this value in the relation for time period, we get:

T = 2 sqrt(0.3219 x12 M / M x 32.2 x 5.5) = 0.9275 seconds