Question

Calculate the rotational inertia of a meter stick, with mass 0.342 kg, about an axis perpendicular to the stick and located at the 22.7 cm mark. (Treat the stick as a thin rod.)

Answer 1

This problem can be solved by applying the parallel axis theorem and noting that the rotational inertia about the midpoint of the stick and perpendicular to the stick is where M is the mass of the stick and L is the length of the stick which is 1 meter here. The rotational inertia about the 22.7 cm mark comes out as 0.054 kg m^2. Detailed solution given below. If you have any doubt regarding the solution, ask me in the comment section. Feedback will be helpful

(ML/12)

solution: (Rotational inertia is also called moment of inertia) Mass of the stick (M) = 0:342 kg Length of the stick (6 = 1.0 mo midpoint of the stick is at u= 0.5 m = 50 cm distance of the axis from midpoint is 150 - 22:7) cm 27.30 m = 27.3 810-2m We will be using the parallel axis theorem which says that moment a inertia about any axis, I is requal to the sum of moment of inertia about an axis passing through the center of mass Io, and Me where i is the distance of the axis from the center of mans axis | Also, Io calculated must, be through a parallel axis. (I= Io + M12). In our case, center of Mass lies at n 250 cm _ (halfway through the rod) I 12 The moment of inertial of the stick through this point about an axis perpendicular to the stick is ML? ,: Joz Me? Тя меня = M(Kre?) 2 0342 x + (27.3X102)) ng m? = 0.342 x 0:157 9. kg m2 0.054 so rotational inertia about 22.7 cm mark is 0.054 kgm .