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Problem 3/155 The light rod is pivoted at O and carries the 2kg and 4kg particles. If the rod is released from rest at $\theta=60°$ and swings in the vertical plane, calculate (a) the velocity $v$ of the 2-kg particle just before it hits the spring in the dashed position and (b) the maximum compression $x$ of the spring. Assume that $x$ is small so that the position of the rod when the spring is compressed is essentially horizontal.

The part I am most confused about it part a) as I am not sure how to come up with the energy equation using angular motion instead of linear motion. Trying to solve the question I assumed I had to use $\frac12I\omega^2$ instead of $\frac12mv^2$.The height for GPE of mass a and mass b I am also unsure how to calculate.

In addition to part a, I assumed the energy equation: $GPE_a=KE_b-GPE_b$ where GPE is gravitational potential energy and KE is kinetic energy. The kinetic energy of A = 0 since it is initially at rest, and GPE of mass b is negative due to loss of GPE where I took the vertical plane as the datum.

I may have the wrong assumptions and I am open to other answers people may have.

Initial energy (all potential):

$$ E_1=g l_1 m_1 \sin (\theta )-g l_2 m_2 \sin (\theta ) $$

Energy just before impact (all kinetic, and neglecting the mass of the rod):

$$ E_2=\frac{1}{2} m_1 \left(l_1 \omega \right){}^2+\frac{1}{2} m_2 \left(l_2 \omega \right){}^2$$

Energy when the spring is completely compressed (all potential, and since $x$ is small neglect the the potential energy from the masses):

$$ E_3=\frac{k x^2}{2} $$

Here $\omega$ is the angular velocity. Solve for $\omega$ from the equation $E_1==E_2$ to get $\omega= 2.58002$. Then the velocity of $m_2$ just before impact can be computed as $v= l_2\omega= 1.16101$.

The compression x can be computed using $E_1==E_3$ and the final equation. Thus $x=\sqrt{\frac{2 E_1}{k}}=0.012$

(Numeric values used $m_1=4,m_2=2,l_1=\frac{300}{1000},l_2=\frac{450}{1000},\theta =60 {}^{\circ},g=9.8,k=35000$)

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