Question

g The mass of a rigid rod is 0.60 kg and its length is 39.0 cm. Calculate the moment of inertia (in kg · m2) for the rod if it is pivoted about an axis through the center of mass of the rod. (Assume the axis is perpendicular to the rod. Enter your answer to at least three significant figures.)

Answer 1

Answer:

0.008 Kg$m^2$

Explanation:

Using the equation of Moment of Inertia about an axis passing through the centre of mass of the rod and perpendicular to it's length

$I_c_m= \frac{1}{12} ML^2$

Where, M is the mass of the rod in Kg and L is the length of the rod in meter.

$I_c_m= \frac{1}{12} \times 0.60\times (0.39)^2$

$I_c_m= 0.008 Kg$ $m^2$