Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A disc of mass m and radius R is attached to a rectangular plate of the same mass breadth R and length 2R . Find moment of inertia of this system about diameter of of disc.They are attached about thier ends. the disc is attached about rectangle length . with explanation pls
Ip=I (com) + Md2 ,p= moment of inertia of axis through p , parallel to centre of mass and at a distance d from centre of mass.Icom = M(R)2/6 for rectangular sheet of breadth R.Therefore , Taking p is the centre of disc and axis parallel to centre of mass of rectangular sheet axis, We get Ip=MR2/6+ M(3R/2)2 {AS DISTANCE BETWEEN COM OF RECTANGLE AND CIRCLE IS 3R/2} Therefore, Ip = 29MR2/12 Adding this to MR2/4 , we get 32MR2/12
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -