TS EAMCET · Physics · Rotational Motion
A particle of mass ' \(\mathrm{m}\) ' is moving along a line \(\mathrm{y}=\mathrm{x}+\mathrm{a}\) with a constant velocity ' \(v\) '. The angular momentum of the particle about the origin is
- A mva
- B \(\mathrm{mva} \sqrt{2}\)
- C \(\frac{\text { mva }}{\sqrt{2}}\)
- D \(\frac{\mathrm{mva}}{\mathrm{x} \sqrt{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\text { mva }}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Given, \(\mathrm{y}=\mathrm{x}+\mathrm{a}\) compairing with \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) \[ \begin{aligned} & \mathrm{m}=1, \mathrm{c}=\mathrm{a} \\ & \tan \theta=1=\tan 45^{\circ} \\ & \theta=45^{\circ} \end{aligned} \] The velocity in vector form…
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