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
2512kg−m2/s
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