Question

two rods of equal mass m and length L lie along the x axis and y axis with their centres at the origin . what is the moment of inertia of both about the line x=y?

Answer 1

Two rods of equal mass $m$ and length $L$ lie along the $x$ axis and $y$ axis with their centers at the origin. What is the moment of inertia of both about the line $x = y$ ?

Solution.

The moment of inertia of both rods about the line $y = x$ :

$$I = 2 I _ {0};$$

$I_0$ - the moment of inertia of one rod about the line $y = x$ .

The mass of the particle of the rod with the length $dx$ :

$$d m = \frac {m}{L} d x;$$

The moment of inertia of the particle of the rod with the length $dx$ about the line $y = x$ :

$$d I _ {0} = d m r ^ {2};$$ $$r ^ {2} = x ^ {2} \sin^ {2} \alpha ;$$ $$d I _ {0} = \frac {m}{L} d x x ^ {2} \sin^ {2} \alpha .$$

The moment of inertia of one rod about the line $y = x$ :

$$I _ {0} = 2 \int_ {0} ^ {\frac {m}{2}} d m r ^ {2} = \frac {2 m}{L} \sin^ {2} \alpha \int_ {0} ^ {\frac {L}{2}} x ^ {2} d x = \frac {2 m}{L} \sin^ {2} (\alpha) \frac {1}{3} \left(\frac {L}{2}\right) ^ {3} = \frac {1}{1 2} m L ^ {2} \sin^ {2} \alpha .$$

The moment of inertia of both rods about the line $y = x$ :

$$I = 2 \frac {1}{1 2} m L ^ {2} \sin^ {2} \alpha = \frac {1}{6} m L ^ {2} \sin^ {2} \alpha .$$

By the diagram:

$$y = x;$$ $$\frac {y}{x} = 1;$$ $$\frac {y}{x} = \tan \alpha ;$$ $$\tan \alpha = 1;$$ $$\alpha = 4 5 {}^ {\circ}.$$ $$I = \frac {1}{6} m L ^ {2} \sin^ {2} (4 5 {}^ {\circ}).$$

Answer: The moment of inertia of both rods about the line $y = x$ is $I = \frac{1}{6} mL^2\sin^2 (45{}^\circ)$ .