Question

Two uniform rods $\text{A}$ and $\text{B}$ are rigidly joined end to end and is pivoted at point $P$, such that it can freely rotate about point $P$ in a vertical plane. A small object moving horizontally, hits the lower end of the combination and stick to it. What should be the velocity of the object so that the system could just be raised to the horizontal position ?

• A. $3.3\text{m/s}$
• B. $4.3\text{m/s}$
• C. $5.3\text{m/s}$
• D. $6.3\text{m/s}$

Answer 1

The correct option is D $6.3\text{m/s}$

Given, $m=0.05\text{kg},m_A=0.01\text{kg},m_B=0.02\text{kg}$ and $l=0.6\text{m}$

The combination of the rods get rotated about point $P$.

So, moment of inertia of the system about the axis passing through point $P$ and perpendicular to the vertical plane (from parallel axis theorem) is given by ,

Moment of inertia of rod $\text{A}$ about $P$ $I_1=\frac{m_A{l}^2}{3}$

Moment of inertia of rod $\text{B}$ about $P$ $I_2=\frac{m_B{l}^2}{12}+m_B{(\frac{l}{2}+l)}^2$

Moment of inertia of object about $P$ $I_3=m(2l{)}^2$

Net moment of inertia about point $P$

$I_p=m(2l{)}^2+\frac{m_A{l}^2}{3}+m_B[\frac{l^2}{12}+{(\frac{l}{2}+l)}^2]$

Substituting the given values, we get

$I_p=0.09{\text{kg-m}}^2$

Since there is no external torque acting on the system about $P$,

From law of conservation of angular momentum we can say that,

$L_i=L_f$ (about $P$)

$\Rightarrow mv\left(2l\right)=I_p\omega$

From the data given in the question,

$0.05\times u\times 2\times 0.6=0.09\times \omega \Rightarrow \omega =0.67u.......\left(1\right)$

Applying law of conservation of mechanical energy after collision

$K_i+U_i\text{(vertical position of rods)}=K_f+U_f\text{(horizontal position of rods)}$

$\Rightarrow \frac{1}{2}{I}_p{\omega }^2+0=0+[mg\left(2l\right)+m_{A}g(\frac{l}{2})+m_{B}g(l+\frac{l}{2})]$

$\Rightarrow \frac{1}{2}\times 0.09\times {\omega }^2=(0.05\times 10\times 2\times 0.6)+(0.01\times 10\times \frac{0.6}{2})+(0.02\times 10\times (0.6+\frac{0.6}{2}))$

$\Rightarrow {\omega }^2=17.64(\text{rad/s}{)}^2\Rightarrow \omega \approx 4.2\text{rad/s}$

Substituting the value of $\omega$ in Eq. $\left(1\right)$, we get

$u=\frac{4.2}{0.67}\approx 6.3\text{m/s}$

Hence, option (d) is the correct answer.