Question

Two uniform rods AB and BC of masses 1 kg and 2 kg respectively having lengths 2 m and 1 m respectively are joined to each other at B. They can rotate freely about B without any friction. The assembly is kept on a smooth horizontal surface as shown in figure.
A horizontal impulse P = 10 Ns is applied on the rod
AB at a distance 0.5 m from point B perpendicular to the rod

Angular momentum of rod BC about centre of rod BC just after application of impulse is

• A.
• B.
• C.
• D. 6 kg - m²/s

Answer 1

The correct option is A

Let's take the part AB

Let $P_1$ be the impulse from the part CB

$P-P_1=m_1{v}_1$

given P = 10 Ns

$10-P_1=I{v}_1$ - - - - - - (1)

If $\omega$ is the angular velocity which AB gains then net angular impulse about the CM of AB (i.e., 0)

= change in angular momentum about 0

$p\times 0.5-p_1\times 1=\frac{m_1{l}_1^2}{12}{\omega }_1$

$5-p_1=\frac{{\omega }_1}{3}$ ...(2)

For the part BC

Applying linear impulse = change in L (momentum)

$P_1=m_2{v}_2\Rightarrow P_1=2v_2$ - - - - - - (3)

Applying angular impulse about 0' = change in angular momentum about 0'

$p_1\times 0.5=\frac{m_1{l}_2^2}{12}{\omega }_2$

$\frac{p_1}{2}=\frac{{\omega }_2}{6}$ ...(4)

Now since end B is common to both rods its velocity must be same as seen from the respective of both rods

$v_1+{\omega }_1\times 1=v_2+{\omega }_2\times 0.5$ - - - - - - (5)

from equations (1), (2), (3), (4) and (5)

$p_1=\frac{25}{6}Ns$

${\omega }_2=12.5rod/s$

$v_1=\frac{35}{6}m/s$

angular momentum of BC about 0'=$\frac{m_1{l}_2^2}{12}{\omega }_2=\frac{2\times 1^2}{12}\times \frac{25}{2}=\frac{25}{12}kg{m}^2/s$