Question

A thin 0.100-kg rod that is 350mm long has a small hole drilled through it 87.5mm from one end. A metal wire is strung through the hole, and the rod is free to rotate about the wire.

Part A

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

Express your answer with the appropriate units.

I = ?

Part B

Determine the period of the rod's oscillation.

Express your answer to two significant digits and include the appropriate units.

T = ?

Answer 1

Moment of Inertia of a thin rod about one end Ie= 1/3 ML^2, where M is mass and L is length of the rod.

Applying parallel axis theorem for moment of inertia, moment of inertia about an axis 87.5 mm from one end is I = Ie + M(87.5)^2 in kg(mm)^2 = 1/3 x 0.1 x (350)^2 + 0.1 x (87.5)^2 = 4083.3 + 765.625 = 4848.925 kg(mm)^2.

The center of gravity is at the center of the rod. Then, the torque acting on the rod = mg x (distance of point of suspension from center).

Center point is at a distance of 175 mm from the end and point of haning is 87.5.

Then torque acting = mg x (distance of center of gravity from the haning point) = 0.1 x 10 (g) x (175 - 87.5) = 1 x 87.5 x 10^-3 Nm = 87.5 x 10^-3 Nm.

At an angle theta from the equilibrium position, torque acting = 87.5 x 10^-3 x theta .

Torque = Moment of Intia x acceleration, 87.5 x 10^-3 x theta = 4.8 x 10^-3 x acceleration,

omega^2 = 87.5/4.8. Then Time period = 2pi/omega = 2 x 3.14 x (4.8/87.5)^1/2 = 6.28 x 0.23 = 1.47 sec.