JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A uniform disc of radius \(R\) and mass \(M\) is free to rotate only about its axis. A string is wrapped over its rim and a body of mass \(m\) is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is
- A \(\frac{{2mg}}{{2m + M}}\)
- B \(\frac{{2Mg}}{{2m + M}}\)
- C \(\frac{{2mg}}{{2M + m}}\)
- D \(\frac{{2Mg}}{{2M + m}}\)
Answer & Solution
Correct Answer
(A) \(\frac{{2mg}}{{2m + M}}\)
Step-by-step Solution
Detailed explanation
From the codition of equilibrium \(mg - T = ma \) \(....(i)\) \(RT =\) \(I\alpha \) \(\begin{array}{l} \overline a = \overline \alpha \times R\,and\,I = \frac{{M{R^2}}}{2}\\ \,\,\,\,\,\,\,\,\,\,\,RT = \frac{{M{R^2}}}{2}.\frac{a}{R} \end{array}\) Tension in string,…
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