TS EAMCET · Physics · Center of Mass Momentum and Collision
A circular disc of radius \(R\) is removed from a bigger circular disc of radius \(2 R\) such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance \(\alpha R\) from the centre of the bigger disc. The value of \(\alpha\) is
- A \(\frac{1}{2}\)
- B \(\frac{1}{3}\)
- C \(\frac{1}{4}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
Let centre \(O_1\) of disc be the origin. Due to symmetry the centre of mass of remaining part will lie at \(x\)-axis. Let \(\mathrm{CM}\) of new disc is at point \(\mathrm{O}_2 \quad\) where, \(\mathrm{O}_1 \mathrm{O}_2=\alpha \mathrm{R}\) (given). here mass of cut off portion,…
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