Question

For the arrangement in the figure, the particle $M_{1}$ attached to one end of a string which moves on a horizontal table in a circle of radius $=\frac{l}{2}$ (where $l$ is the length of the string) with the constant angular speed $\omega$ . The other end of the string attached to mass $M_{2}$ which rests on a vertical rod. When the rod collapse, the acceleration of mass $M_{2}$ at that instant

• A. $g$
• B. $\frac{\omega ^{2} l}{2}$
• C. $\frac{2 M_{2} g - M_{1} l \left(\omega \right)^{2}}{2 \left(M_{1} + M_{2}\right)}$
• D. $\frac{M_{2} g + M_{1} l \omega ^{2}}{M_{1} + M_{2}}$

Answer 1

Answer: (C)

$\frac{2 M_{2} g - M_{1} l \left(\omega \right)^{2}}{2 \left(M_{1} + M_{2}\right)}$

Answer 2

From Newton's second law $T-M_{1}\omega ^{2}\frac{l}{2}=M_{1}a$ And $M_{2} \, g-T=M_{2}a$ After solving above equations, we get $a=\frac{2 M_{2} g - M_{1} l \left(\omega \right)^{2}}{2 \left(M_{1} + M_{2}\right)}$