A bar of length $L$ and mass $M$ has a frictionless pivot through its mid point. An additional small block of mass $2 M$ is attached on the right end of the bar and small block of mass $M$ is attached on the left end of the bar. The bar is held in horizontal position by a vertical cord attached to a point at distance L/4 from the left end as shown in the figure. When the bar has been rotated through $90^{o}$ and is vertical, what is the speed of the mass 2M?

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Solution

$Δ U$ In Process $= - 2 m \frac{L}{2} g + m \frac{L}{2} g = - m \frac{L}{2} g$
$Δ U + Δ K = 0 \Rightarrow Δ K = - Δ U$
$\Rightarrow Δ K = m \frac{L}{2} g$
$\frac{1}{2} I ω^{2} = m \frac{L}{2} g \Rightarrow \frac{1}{2} . \frac{5}{6} m L^{2} ω^{2} = m \frac{L}{2} g \Rightarrow ω = \sqrt{\frac{6 g}{5 L}}$
Velocity of $2 m = \frac{L}{2} ω = \sqrt{\frac{3 g L}{k}}$

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ΔU In Process =−2mL2g+mL2g=−mL2gΔU+ΔK=0⇒ΔK=−ΔU⇒ΔK=mL2g12Iω2=mL2g⇒12.56mL2ω2=mL2g⇒ω=√6g5LVelocity of 2m=L2ω=√3gLk