JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A bead of mass \(m\) stays at point \(P ( a , b )\) on a wire bent in the shape of a parabola \(y=4 Cx ^{2}\) and rotating with angular speed \(\omega\) (see figure). The value of \(\omega\) is (neglect friction)
- A \(\sqrt{\frac{2 gC }{ ab }}\)
- B \(2 \sqrt{2 g C}\)
- C \(\sqrt{\frac{2 g}{C}}\)
- D \(2 \sqrt{ gC }\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{2 g C}\)
Step-by-step Solution
Detailed explanation
In rotating frame \(mx \omega^{2} \cos \theta= mg \sin \theta\) \(x \omega^{2}= g \tan \theta\) \(x \omega^{2}= g \cdot \frac{ dy }{ dx }\) \(x \omega^{2}= g \cdot(8 cx )\) \(\omega^{2}=8 gc\) \(\omega=2 \sqrt{2 gc }\)
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