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A uniform circular disk of radius R and mass M is rotating with angular speed w about an axis, passing through its center and inclined at an angle 60 degrees with respect to its symmetry axis. The magnitude of the angular momentum of the disk is,

🗓 Oct 14, 2024

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Anonymous 🌐 Kerala
L =I(X)w+I(Y)w+I(Z)w
Magnintude of L =sq.root (I(X)w)^2+(I(Y)w)^2+(I(Z)w)^2)
I(X) =MR^2/4=I(Y)
I(Z)=MR^2/2
W(x)=wsin60°
W(y) =0
W(z)=wcos60°
When we apply these values in formula
We get after simplification
Sq. Root(((M^2×R^4×w^2)/16))×(3/4 +1))
= (MR^2w/4)(sq.root 7/2)
=(Root 7 MR^2w)/8
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Anonymous 🌐 India
Wrong Answer
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