Question

A thin 0.200-kg rod that is 350 mm long has a small hole drilled through it 87.5 mm from one end. A metal wire is strung through the hole, and the rod is free to rotate about the wire.

A. Determine the rod's rotational inertia I about this axis.

B. Determine the period of the rod's oscillation.

Answer 1

A.

The rotational inertia of the rod about its center of mass is

I₀ = ML² / 12

Here, mass of the rod is M and length of the rod is L.

The diagram representing the wire and the thin rod is shown below:

Using parallel axes theorem, the rotational inertia about the point where the metal wire strung is

I = I₀ + Mh²

   = ML² / 12 + Mh²

Substitute 0.200 kg for M, 350 mm for L and 87.5 m for h in the above equation,

I = ML² / 12 + Mh²

= (0.200 kg)( 350 mm)² (1 m / 1000 mm)² / 12 + (0.200 kg) (87.5 mm)² (1 m / 1000 mm)²

= 2041.7 x 10^-6 kg m² + 1531.25 x 10^-6 kg m²

= 3.5729 x 10^-3 kg m²

Rounding off to three significant figures, the rotational inertia of the thin rod about the axis where the metal wire strung is 3.57 x 10^-3 kg m².

B.

The rotation of the rod about the axis where the metal wire strung is same as the swinging of physical pendulum having pivoted at the point A.

Now, the period of the rotating rod is

Here, acceleration due to gravity is g.

Substitute 3.57 x 10^-3 kg m² for I, 0.200 kg for M, 9.8 m/s² for g and 87.5 m for h in the above equation,

Rounding off to three significant figures, the period of the rotation of the rod is 0.0286 s.