AP EAMCET · PHYSICS · Center of Mass Momentum and Collision
A circular portion of radius \(R_2\) has been removed from one edge of a circular disc of radius \(R_1\). The correct expression for the centre of mass for the remaining portion of the disc is
- A \(-\frac{R_2^2}{R_1+R_2}\)
- B \(-\frac{R_2^2}{R_1-R_2}\)
- C \(\frac{R_2^2}{R_1+R_2}\)
- D \(-\frac{R_1^2}{R_1+R_2}\)
Answer & Solution
Correct Answer
(A) \(-\frac{R_2^2}{R_1+R_2}\)
Step-by-step Solution
Detailed explanation
Given situation is shown below. Here we treat removed mass as negative mass. As solid is symmetrical about \(X\) - axis, its CM lies along \(X\) - axis. Now \(x_{\mathrm{CM}}=\frac{m_1 x_1-m_2 x_2}{m_1-m_2}\)…
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