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Why is the final kinetic energy of the particle of mass m zero.It should have some velocity when it reaches the bottom??

Can we solve this solve by using the concept of reduced mass system?

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  • $\begingroup$ Typeset mathematical terms using MathJax rather than posting image of question. Here's the tutorial. $\endgroup$ – SarGe Jul 8 at 5:25
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Because we need the minimum speed by which the circle gets completed i.e. it stops when the circle is just completed. If the speed given to $m$ is greater than $\displaystyle \sqrt{\frac{4gl}{3}}$ then it completes the circle as well as have some velocity at the lowest point.

The torque of mass $2m$ is greater than that of $m$ about the center of rod and hence it will tend to rotate the rod on clockwise direction (if velocity given to $m$ is towards left). But, if $m$ has sufficient angular momentum, then it can it can complete the circle. However, the torque due to $2m$ will oppose this motion, too.

Also, for a system of two particles with masses $m_1$ and $m_2$ exerting equal and opposite forces on each other and subject to no external forces, concept of reduced mass can be used which is not the case here.

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  • $\begingroup$ Can we solve it by using the concept of reduced mass $\endgroup$ – NASA Jul 8 at 5:34
  • $\begingroup$ but how will the mass m stop at the bottom.It must have some velocity $\endgroup$ – NASA Jul 8 at 5:35
  • $\begingroup$ the mass 2m will have final velocity equal to zero when it reaches the top and mass m would be having some velocity when it reaches the bottom $\endgroup$ – NASA Jul 8 at 5:46
  • $\begingroup$ No, both will be at rest for $v_{min}$. $\endgroup$ – SarGe Jul 8 at 5:50
  • $\begingroup$ okay so can we solve this question by using the concept of reduced mass $\endgroup$ – NASA Jul 8 at 6:12

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