Question

A metre stick weighing 240 g is pivoted at its upped end in such a way that it can freely rotate in a vertical plane through this end (figure 10-E12). A particle of mass 100 g is attached to the upped end of the stick through a light string of length 1 m. Initially, the rod is kept vertical and the string horizontal when the system is released from rest. The particle collides with the lower end of the stick and sticks there. Find the maximum angle through which the stick will rise.

Answer 1

$\frac{1}{2}I{\omega }^2-0=0.1\times 10\times 1$

$\Rightarrow \omega =\sqrt{20}$

For, collision $0.1\times 12\times \sqrt{20}+0$

= $\left[(\frac{0.24}{3}\times 12+(0.1{)}^2\left(1{)}^2\right)\right]\omega$

$\Rightarrow \omega =\frac{\sqrt{20}}{[10\times (0.18\left)\right]}$

$\Rightarrow 0-\frac{1}{2}I{\omega }^2=-m_{1}gI(1-cos\theta )$

$-m_{2}g\frac{1}{2}(1-cos\theta )$

$-0.24\times 10\times 0.5(1-cos\theta )$

$\Rightarrow \frac{1}{2}\times 0.18\times \left(\frac{20}{324}\right)=2.2\times 9(1-cos\theta )$

$\Rightarrow (1-cos\theta )=\frac{1}{(2.2\times 1.8)}$

$\Rightarrow 1-cos\theta =0.252$

$\Rightarrow cos\theta =1-0.252=0.748$

$\Rightarrow W=co{s}^{-1}\left(0.748\right)={41}^{\circ }$