Answer to Question #321340 in Mechanics | Relativity for Quân Jason

Question

Calculate the rotational inertia of a meter stick, with mass 0.24 kg, about an axis perpendicular to the stick and located at the 30 cm mark. (Treat the stick as a third rod.)

a) 90.7 x 10^-3 kg.m²

b) 41.6 x 10^-3 kg.m²

c) 20 x 10^-3 kg.m²

d) 29.6 x 10^-3 kg.m²

Answer 1

$$I=I_c+md^2=\frac{1}{12}ml^2+md^2$$

$$I=\frac{1}{12}*0.24*1.0^2+0.24*0.2^2=29.6*10^{-3}\:\rm kg\cdot m^2$$

Answer: d