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JEE Advanced · Physics · 8. Rotational Motion

A rod of mass m and length L, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets embedded in it. The combined system now rotates with an angular speed ω about the pivot. The maximum angular speed ωM is achieved for x=xM. Then

  1. A ω=3vxL2+3x2
  2. B ω=12vxL2+12x2
  3. C xM=L3
  4. D ωM=v2L3
Verified Solution

Answer & Solution

Correct Answer

(A) ω=3vxL2+3x2

Step-by-step Solution

Detailed explanation

The net torque on the system (rod+bullet) will be zero about hinge point. Therefore, we can apply the principle of angular momentum conservation about hinge point.
From angular momentum conservation,
\(m v x=\left(\frac{m L^2}{3}+m x^2\right) \omega\)
\(\Rightarrow \omega=\frac{3 v x}{L^2+3 x^2}=\frac{3 v}{\left(\frac{L^2}{x}+3 x\right)}\)
For maximum angular velocity,
\(\frac{ d \omega}{ d x}=0\)
\(\Rightarrow \frac{ d }{ d x}\left(\frac{L^2}{x}+3 x\right)=0 \Rightarrow \frac{-L^2}{x^2}+3=0 \Rightarrow\) \(x=\frac{L}{\sqrt{3}}\)
\(\Rightarrow \omega_{\max }=\frac{3 v}{\left(\frac{\sqrt{3} L^2}{L}+\sqrt{3} L\right)}=\frac{\sqrt{3} v }{2 L}\)
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