solution
ifF=300N
r=2.5m
m=10kg
u=0m/s
t=8s
fromF=ma
300=10∗a
a=300/10
a=30m/s2
fromv=u+at
v=0+30∗8
v=240m/s
linearmomentum p=m∗v.
=10∗240
=2400kgm/s
the final angular momentum L = distance of the object from a rotation axis(r) ∗ the linear momentum(p)
L=r∗p
=2.5m∗2400kgm2/s
=6000kgm2/s
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