Question

A light rod carries three equal masses A , B and C as shown in the figure. What will be the velocity of B in the vertical position of the rod, if it is released from the horizontal position as shown in the figure

• A. $\sqrt{\frac{8g\ell }{7}}$
• B. $\sqrt{\frac{4g\ell }{7}}$
• C. $\sqrt{\frac{2g\ell }{7}}$
• D. $\sqrt{\frac{10g\ell }{7}}$

Answer 1

The correct option is A $\sqrt{\frac{8g\ell }{7}}$

$\begin{array}{c}Accordingtoquestion.....................\\ I=Mofrodaboutfixedpoint\\ I=M{\left(\frac{l}{3}\right)}^2+M{\left[\frac{2l}{3}\right]}^2+M{l}^2\\ =M{l}^2\left[\left(\frac{1}{9}\right)+\left(\frac{4}{9}\right)+1\right]\\ =M{l}^2\left[\frac{14}{9}\right]\\ \begin{array}{c}=\left[\frac{14}{9}\right]M{l}^2-----------\left(i\right)\\ Now,\\ whenrodisreleased,\\ lossinPE=gaininKE\\ \begin{array}{c}\\ \Rightarrow 2Mgl=\left(\frac{1}{2}\right)\times \left[\frac{14}{9}\right]M{l}^2{\omega }^2-----from\left(1\right)\\ \begin{array}{c}\Rightarrow 2g=\left(\frac{14}{18}\right)l{\omega }^2\\ Hence,\\ \Rightarrow {\omega }^2=\left[\frac{36g}{14l}\right]\\ \therefore V=r\omega \\ Now,\\ V_B=\left(\frac{2}{3}\right)l\times \sqrt{\left[\frac{36g}{14l}\right]}\\ \Rightarrow \sqrt{\left(\frac{36g}{14l}\right)\times \left(\frac{4}{9}\right)l^2}\end{array}\\ \Rightarrow \sqrt{l\times g\times \frac{16}{14}}=\sqrt{\left(\frac{8}{7}\right)gl}\\ \therefore \sqrt{\frac{8gl}{7}}\\ sothatthecorrectoptionisA.\end{array}\end{array}\end{array}$