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  • A rod of length '  ', having mass ' <img alt="m" src="https://entrancecorner.oncodecogs.com/gif.latex?m"

A rod of length ' L ', having mass ' m ' and ends ' X ' and ' Y ' is lying on a horizontal plane. A particle of mass ' m ' attached to end ' X ' with a massless string of length ' L ', equal to the length of the rod.


If the mass of the particle is also ' m ' and it is given velocity ' v ' perpendicular to the rod. Then the acceleration of the centre of mass, just after projecting the particle will be ______

Option: 1

\frac{m v^2}{5 L}


Option: 2

\frac{2m v^2}{5 L}


Option: 3

\frac{m v^2}{15 L}


Option: 4

\frac{m v^2}{7 L}


Answers (1)

best_answer

a_x=a_{C.M.}+\left(\frac{L}{2}\right) \alpha

For the rod:

\begin{aligned} T&=m a_{C.M.} \\ T\left(\frac{L}{2}\right)&=\left(\frac{1}{12} m L^2\right) \alpha\\T&=\frac{m L^2 \alpha}{12} \times \frac{2}{L}=\left(\frac{m L}{6}\right) \alpha \end{aligned}

Observing the motion of the particle from end '  '


As, the particle is performing circular motion about the end ' $x$ '.

\begin{aligned} T+m a_x&=\frac{m v^2}{L} \\ T+m a_{C\cdot M.}+m\left(\frac{L}{2}\right)\alpha&=\frac{m v^2}{L} \\ T+T+3 T&=\frac{m v^2}{L} \\ 5 T&=\frac{m v^2}{L} \\ T&=\frac{m v^2}{5 L} \end{aligned}

Posted by

Kuldeep Maurya

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