Question

A metre stick weighing 240 g is pivoted at its upper end in such a way that it can freely rotate in a vertical place through this end (figure 10-E12). A particle of mass 100 g is attached to the upper 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.

Figure

Answer 1

Let the angular velocity of stick after the collision be ω.
Given:
Mass of the stick = m = 240 g
Mass of the particle = m' = 100 g
Length of the string (or rod) = r = l = 1 m
Moment of inertia of the particle about the pivoted end = $I\text{'}=m\text{'}{r}^2$
$$\begin{gathered} \Rightarrow I\text{'}=\left(0.1\times 1^2\right) \\ \Rightarrow I\text{'}=0.1\mathrm{kg}-{\mathrm{m}}^2 \end{gathered}$$
Moment of inertia of the rod about the pivoted end = $I=\frac{m{l}^2}{3}$
$$\begin{gathered} \Rightarrow I=\left(\frac{0.24}{3}\times 1^2\right) \\ \Rightarrow I=0.08\mathrm{kg}-{\mathrm{m}}^2 \end{gathered}$$

Applying the law of conservation of energy, we get:
Final energy of the particle = Initial energy of the rod
$$\begin{gathered} mgh=\frac{1}{2}I{\omega }^2 \\ \Rightarrow \frac{1}{2}I{\omega }^2=0.1\times 10\times 1 \\ \Rightarrow \omega =\sqrt{20} \end{gathered}$$

For collision, we have:
$0.1\times 1^2\times \sqrt{20}+0$
$$\begin{gathered} =\left[\left(\frac{0.24}{3}\right)\times 1^2+{\left(0.1\right)}^2{\left(1\right)}^2\right]\omega \\ \Rightarrow \omega =\frac{\sqrt{20}}{\left[10\times \left(0.18\right)\right]} \\ \Rightarrow 0-\frac{1}{2}I{\omega }^2=-m_{1}gl\left(1-\cos \mathrm{\theta }\right)-m_{2}g\frac{1}{2}\left(1-\cos \mathrm{\theta }\right) \\ \Rightarrow \frac{1}{2}\mathit{}I{\omega }^2=-mgl\left(1-\cos \theta \right)-m_{2}g\frac{1}{2}\left(1-\cos \mathrm{\theta }\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{2}I{\omega }^2=0.1\times 10\left(1-\cos \mathrm{\theta }\right)-0.24\times 10\times 0.5\left(1-\cos \mathrm{\theta }\right) \\ \Rightarrow \frac{1}{2}\times 0.18\times \left(\frac{20}{324}\right)=2.2\times 9\left(1-\cos \mathrm{\theta }\right) \\ \Rightarrow \left(1-\cos \mathrm{\theta }\right)=\frac{1}{\left(2.2\times 1.8\right)} \\ \Rightarrow 1-\cos \mathrm{\theta }=0.252 \\ \Rightarrow \cos \mathrm{\theta }=1-0.252=0.748 \\ \Rightarrow \theta ={\cos }^{-1}\left(0.748\right)=41° \end{gathered}$$