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MatinSAR

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- Homework Statement
- A rod of mass M and length L is rotating freely around an axis at one end. There are two particles attached to the rod at distances of L/3 and 2L/3 from the axis, with masses m and 2m, respectively. The system is only affected by gravitational forces.

A) Determine the angular acceleration of the rod and particles under these conditions. B) Calculate the angular velocity of the system when the rod is vertical after rotating 90 degrees.

- Relevant Equations
- Rigid body dynamics.

For the rod and the two particles ##x_1## is not zero but we have:$$x_2=x_3=0$$ We start with inertia tensor of the rod: $$ I_{ij} = \int \lambda ( \delta _{ij} \sum_{k} x^2_k -x_i x_j ) dx_1$$ $$ \lambda = constant = M/L $$ Using above equations, we have:

$$I_{11} = \dfrac M L \int ( x^2_2 +x^2_3 ) dx_1=0$$ $$I_{22} = \dfrac M L \int ( x^2_1 +x^2_3 ) dx_1=\dfrac {1}{3} ML^2$$ $$I_{33} = \dfrac M L \int ( x^2_1 +x^2_2 ) dx_1=\dfrac {1}{3} ML^2$$ $$I_{12}=...=I_{23}=...=0$$

For the two particles: $$I_{ij}=m_1(\delta _{ij} \sum_{k} x^2_{1k} -x_{1i} x_{1j})+m_2(\delta _{ij} \sum_{k} x^2_{2k} -x_{2i} x_{2j})$$ After Substituting from the information provided in problem, we get: $$I_{11}=m_1(x^2_{12}+x^2_{13})+m_2(x^2_{22}+x^2_{23}) = 0$$ $$I_{22}=m_1(x^2_{11}+x^2_{13})+m_2(x^2_{21}+x^2_{23}) =mL^2$$ $$I_{33}=m_1(x^2_{11}+x^2_{12})+m_2(x^2_{21}+x^2_{22}) =mL^2$$$$I_{12}=...=I_{23}=...=0$$

Therefore, the rigid body has an inertia tensor given by: $$I_{1}=0$$$$I_{2}=I_{3}=(\dfrac M 3 +m)L^2=\beta$$ Euler's equations becomes : $$ - \omega_1 \omega_2- \dot \omega_3=N_3 $$ $$\omega_1 \omega_3- \dot \omega_2=N_2$$ $$\dot \omega_1=N_1 $$

Could you please verify if the steps I have taken so far are correct? Any guidance or feedback you can provide would be very helpful.

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