Question

Two uniform rods A and B of length 0.6 m each and of masses 1 kg and 2 kg respectively are rigidly joined end to end. The combination is pivoted at the lighter end P as shown in figure such that it can freely rotate about point P in a vertical plane. A small object of mass 5 kg, moving horizontally, hits the lower end of the combination and sticks to it. What should be the velocity (in m/s) of the object, so that the system could just be raised to the horizontal position. (Take $g = 9.8 m/s^2$)

Answer 1

Velocity required: $v\approx6.3\,\mathrm{m/s}$.

Answer 2

Solution:

1. Moments of Inertia:

   • Rod A (1 kg, length 0.6 m; pivot at one end):

     $$I_A=\frac{1}{3}mL^2=\frac{1}{3}\times1\times(0.6)^2=0.12\;\mathrm{kg\,m^2}$$

   • Rod B (2 kg, length 0.6 m):
     Its center of mass is at $0.6+0.3=0.9\;\mathrm{m}$ from the pivot. Using the parallel axis theorem,

     $$I_{B,\text{cm}}=\frac{1}{12}mL^2=\frac{1}{12}\times2\times(0.6)^2=0.06\;\mathrm{kg\,m^2}$$

     and

     $$I_B=I_{B,\text{cm}}+m(0.9)^2=0.06+2\times0.81=1.68\;\mathrm{kg\,m^2}.$$

   • Object (5 kg) attached at the lower end (1.2 m from pivot):

     $$I_{\text{obj}}=5\times(1.2)^2=7.2\;\mathrm{kg\,m^2}.$$

   • Total Moment of Inertia:

     $$I_{\text{total}}=I_A+I_B+I_{\text{obj}}=0.12+1.68+7.2=9\;\mathrm{kg\,m^2}.$$

2. Conservation of Angular Momentum at Collision:

   The object (mass $5$ kg) moving horizontally with speed $v$ has angular momentum (about the pivot)

   $$L_{\text{initial}}=5v\times1.2=6v.$$

   After the collision, the combined system rotates with angular speed $\omega$ so that

   $$I_{\text{total}}\omega=6v \quad\Rightarrow\quad \omega=\frac{6v}{9}=\frac{2v}{3}.$$

3. Energy Conservation (Swing up to Horizontal):

   As the system rotates from vertical (initial) to horizontal:

   • Rise in potential energy for each mass:

     For any mass at distance $r$ from the pivot, the rise is $r$, because initially it is at $-r$ (if we choose the pivot as reference) and becomes at 0 in the horizontal position.

     • Rod A: $r=0.3\,\mathrm{m}$
       $\Delta PE_A= 1\times9.8\times0.3=2.94\,\mathrm{J}$.
     • Rod B: $r=0.9\,\mathrm{m}$
       $\Delta PE_B= 2\times9.8\times0.9=17.64\,\mathrm{J}$.
     • Object: $r=1.2\,\mathrm{m}$
       $\Delta PE_{\text{obj}}=5\times9.8\times1.2=58.8\,\mathrm{J}$.

   • Total increase in potential energy:

     $$\Delta PE=2.94+17.64+58.8=79.38\,\mathrm{J}.$$

   The initial kinetic energy right after collision is:

   $$KE_0=\frac{1}{2}I_{\text{total}}\omega^2=\frac{1}{2}\times9\times\left(\frac{2v}{3}\right)^2=\frac{1}{2}\times9\times\frac{4v^2}{9}=2v^2.$$

   Setting $KE_0=\Delta PE$:

   $$2v^2=79.38\quad\Rightarrow\quad v^2=39.69\quad\Rightarrow\quad v=\sqrt{39.69}\approx6.3\,\mathrm{m/s}.$$

---

Core Explanation:

1. Compute individual moments of inertia of rods and object about the pivot.
2. Use angular momentum conservation to relate $v$ and $\omega$.
3. Equate the rotational kinetic energy to the gain in gravitational potential energy (using the rise $r$ for each mass) to solve for $v$.