A rectangular block, with face lengths a = 33.0 cm and b = 45.0 cm, is to be sus
ID: 1491286 • Letter: A
Question
A rectangular block, with face lengths a = 33.0 cm and b = 45.0 cm, is to be suspended on a thin horizontal rod running through a narrow hole in the block. The block is then to be set swinging about the rod like a pendulum, through small angles so that it is in SHM. The figure shows one possible position of the hole, at distance r from the block's center, along a line connecting the center with a corner, For what value of r does the minimum the period of the pendulum occur? There is actually a line of points around the block's center for which the period of swinging has the same minimum value, What shape does that line make?Explanation / Answer
The rotational inertia of the block around the rotational axis is obtained by the parallel axis theorem:
I = ICOM + mr2 = (m/2)(a2 + b2 ) + mr2
The period of the oscillation is T(r) = 2(I/mgr) = 2[(a2 + b2 )/(12gr) + r/g]
When the period takes the minimum value, the derivative of the period is zero. For convenience, we consider the function inside the square root, take the derivative of the function and set it equal to zero.
For if this function takes a minimum value, the period takes a minimum value too.
f(r) = (a2 + b2 )/(12gr) + r/g,
f’(r) = -(a2 + b2 )/(12gr2 ) + 1/g = 0
rmin = [(a2 + b2 )/(12)] = [(0.332 + 0.452 )/(12)] = 0.1610 m............Ans.
(b) Plugging the result of part a) into the expression for the period, we get
T(rmin) = 2[(a2 + b2 )/(12gr) + rmin/g] = 0.823 s