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,

Question

Anonymous · Accepted Answer

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

🤔 No result found

👉 Don't loose hope, Try out below methods

Feedback ( Website kaise lagi? 🤔)

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,

Comments and Discussions

Was your question answered?

Want to get famous? 📸

Answer question and get 🤩 high-lighted on our homepage.

People Also Viewed