Answer:
(a). The total moment of inertia of the system is 0.0233 kg-m².
(b). The kinetic energy is 0.0011 J.
Explanation:
Given that,
Mass of each particle m= 0.85 kg
Length = 5.6 cm
Mass of each rod M= 1.2 kg
Angular speed = 0.30 rad/s
The moment of inertia of the rod between axis of rotation and mass is
![I_{1}=\dfrac{Md^2}{3}](https://tex.z-dn.net/?f=I_%7B1%7D%3D%5Cdfrac%7BMd%5E2%7D%7B3%7D)
The moment of inertia of the rod between masses about center of mass is
![I_{cm}=\dfrac{Md^2}{12}](https://tex.z-dn.net/?f=I_%7Bcm%7D%3D%5Cdfrac%7BMd%5E2%7D%7B12%7D)
Moment of inertial of the rod between masses about point O is
![I_{2}=M(d+\dfrac{d}{2})^2+\dfrac{Md^2}{12}](https://tex.z-dn.net/?f=I_%7B2%7D%3DM%28d%2B%5Cdfrac%7Bd%7D%7B2%7D%29%5E2%2B%5Cdfrac%7BMd%5E2%7D%7B12%7D)
Moment of inertia of two masses is
![I_{m}=md^2+m(2d)^2](https://tex.z-dn.net/?f=I_%7Bm%7D%3Dmd%5E2%2Bm%282d%29%5E2)
(a). We need to calculate the total moment of inertia of the system
Using formula of moment of inertia
![I_{t}=I_{1}+I_{2}+I_{m}](https://tex.z-dn.net/?f=I_%7Bt%7D%3DI_%7B1%7D%2BI_%7B2%7D%2BI_%7Bm%7D)
Put the value into the formula
![I_{t}=\dfrac{Md^2}{3}+M(d+\dfrac{d}{2})^2+\dfrac{Md^2}{12}+md^2+m(2d)^2](https://tex.z-dn.net/?f=I_%7Bt%7D%3D%5Cdfrac%7BMd%5E2%7D%7B3%7D%2BM%28d%2B%5Cdfrac%7Bd%7D%7B2%7D%29%5E2%2B%5Cdfrac%7BMd%5E2%7D%7B12%7D%2Bmd%5E2%2Bm%282d%29%5E2)
![I_{t}=\dfrac{Md^2}{3}+\dfrac{9Md^2}{4}+\dfrac{Md^2}{12}+md^2+4md^2](https://tex.z-dn.net/?f=I_%7Bt%7D%3D%5Cdfrac%7BMd%5E2%7D%7B3%7D%2B%5Cdfrac%7B9Md%5E2%7D%7B4%7D%2B%5Cdfrac%7BMd%5E2%7D%7B12%7D%2Bmd%5E2%2B4md%5E2)
![I_{t}=\dfrac{32Md^2}{12}+5md^2](https://tex.z-dn.net/?f=I_%7Bt%7D%3D%5Cdfrac%7B32Md%5E2%7D%7B12%7D%2B5md%5E2)
![I_{t}=2.67Md^2+md^2](https://tex.z-dn.net/?f=I_%7Bt%7D%3D2.67Md%5E2%2Bmd%5E2)
Put the value into the formula
![I_{t}=2.67\times1.2\times(5.6\times10^{-2})^2+5\times0.85\times(5.6\times10^{-2})^2](https://tex.z-dn.net/?f=I_%7Bt%7D%3D2.67%5Ctimes1.2%5Ctimes%285.6%5Ctimes10%5E%7B-2%7D%29%5E2%2B5%5Ctimes0.85%5Ctimes%285.6%5Ctimes10%5E%7B-2%7D%29%5E2)
![I_{t}=0.0233\ kg-m^{2}](https://tex.z-dn.net/?f=I_%7Bt%7D%3D0.0233%5C%20kg-m%5E%7B2%7D)
(b). We need to calculate the kinetic energy
Using formula of kinetic energy
![K.E=\dfrac{1}{2}I\omega^2](https://tex.z-dn.net/?f=K.E%3D%5Cdfrac%7B1%7D%7B2%7DI%5Comega%5E2)
Put the value into the formula
![K.E=\dfrac{1}{2}\times0.0233\times(0.30)^2](https://tex.z-dn.net/?f=K.E%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes0.0233%5Ctimes%280.30%29%5E2)
![K.E=0.0011\ J](https://tex.z-dn.net/?f=K.E%3D0.0011%5C%20J)
Hence, (a). The total moment of inertia of the system is 0.0233 kg-m².
(b). The kinetic energy is 0.0011 J.