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
Let's take the part AB
Let P1 be the impulse from the part CB
P−P1=m1v1
given P = 10 Ns
10−P1=Iv1 - - - - - - (1)
If ω 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×0.5−p1×1=m1l2112ω1
5−p1=ω13 ...(2)
For the part BC
Applying linear impulse = change in L (momentum)
P1=m2v2⇒P1=2v2 - - - - - - (3)
Applying angular impulse about 0' = change in angular momentum about 0'
p1×0.5=m1l2212ω2
p12=ω26 ...(4)
Now since end B is common to both rods its velocity must be same as seen from the respective of both rods
v1+ω1×1=v2+ω2×0.5 - - - - - - (5)
from equations (1), (2), (3), (4) and (5)
p1=256Ns
ω2=12.5rod/s
v1=356m/s
angular momentum of BC about 0'=m1l2212ω2=2×1212×252=2512kgm2/s