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Answer 1

Answer :

A::C

Solution :

(a,c) in ΔOAM,OM=12+a2−−−−−−√=2ha2+a2−−−−−−−−√=5a

The circumference of a circle of radius OM will be `2pi(5a)

=10 pia. For completing this circle once, the smaller disc will

have to take 10πa2πa=5rounds.

Therefore the C.M. of the assembly rotates about z-

axis with an angular speed of w/5.

The angular momentum about the C.M. of the system

Lc=Icw=[12ma2]ω

+[12×4m×(2a)2]×ω=17ma2ω2

Now vc=m×ωa+4m×2oemga5m=9ωa5

and r⊥=ml+4m×2I5m9I5

LofC.M.=5m×9ωa5×9I5=81mωa2×24−−√5

Lz=81mωa224−−√5cosθ−Icωsinθ

=81mω2245−−−√×245−−−√−17ma2ω10)=113450mωa2.

Answer 2

Answer:

why do we multiply 6,3,2 here