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The Principle

On a horizont...

Question

On a horizontal smooth plane, a particle collides with the rod hinged at one point(as shown in the figure) and sticks to it. Find the final angular velocity of the system.

$\frac{3 m_{2} u l}{m_{2} + 3 m_{1}}$

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$\frac{3 m_{1} u}{(m_{2} + 3 m_{1}) l}$

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$\frac{3 m_{1} u}{(m_{2} + 3 m_{1}) l}$

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$\frac{2 m_{1} u l}{3 (m_{2} + 3 m_{1})}$

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Solution

The correct option is B
$\frac{3 m_{1} u}{(m_{2} + 3 m_{1}) l}$

(a) $τ_{H} = 0$ see L about hinge is conserved initially $L_{H} = m_{1} u l$

finally $L_{H} = (\frac{m_{2} l^{2}}{3} + m_{1} l^{2}) ω \Rightarrow ω = \frac{3 m_{1} u}{(m_{2} + 3 m_{1}) l}$

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On a horizontal smooth plane, a particle collides with the rod hinged at one point(as shown in the figure) and sticks to it. Find the final angular velocity of the system.

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On a horizontal smooth plane, a particle collides with the rod hinged at one point(as shown in the figure) and sticks to it. Find the final angular velocity of the system.

The correct option is B 3m1u(m2+3m1)l (a) τH=0 see L about hinge is conserved initially LH=m1ul finally LH=(m2l23+m1l2)ω⇒ ω = 3m1u(m2+3m1)l

The correct option is B
$\frac{3 m_{1} u}{(m_{2} + 3 m_{1}) l}$

(a) $τ_{H} = 0$ see L about hinge is conserved initially $L_{H} = m_{1} u l$

finally $L_{H} = (\frac{m_{2} l^{2}}{3} + m_{1} l^{2}) ω \Rightarrow ω = \frac{3 m_{1} u}{(m_{2} + 3 m_{1}) l}$