Moment of Inertia Word Scramble
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| Question | Answer |
| Quarter-Circular Arc | Centroid (x,y)= (2r/π, 2r/π) MOI= πr/2 |
| Semi-Circular Arc | Centroid (x,y)= (0, 2r/π) MOI= πr |
| Arc of Circle | Centroid (x,y)= (rsinα/α,0) MOI= 2αr |
| Triangular Area | Centroid (x,y)= (0,h/3) MOI= bh/3 |
| Quarter circular area | Centroid (x,y)= (4r/3π, 4r/3π) MOI= πr²/2 |
| Quarter Elliptical Area | Centroid (x,y)= (4a/3π, 4b/3π) MOI= πab/4 |
| Semi Elliptical Area | Centroid (x,y)= (0, 4b/3π) MOI= πab/2 |
| Parabolic spandrel (y=kx²) | Centroid (x,y)= (3a/4, 3h/10) MOI= ah/3 |
| General spandrel (y=kx^n) | Centroid (x,y)= ([n+1)/(n+2)]a,[(n+1)/(4n+2)]h) MOI= ah/n+1 |
| Circular Sector | Centroid (x,y)= (2rsinα/3α,0) MOI= αr² |
| Rectangle | MOI about centroidal axis(Ix'x',Iy'y')= bh³/12, b³h/12 MOI about axes (Ixx,Iyy)= bh³/3, b³h/3 Polar moment of inertia= Ixx+Iyy |
| Triangle | Ix'x'=bh³/36 Ixx=bh³/12 |
| Circle | Ixx= Iyy= πR^4/4 Jo= πR^4/2 |
| Semicircle | Ixx= Iyy= πR64/8 Jo= πR^4/2 Ix'x'= 0.11R64 |
| Quarter Circle | Ixx= Iyy= πR^4/16 Jo= πR^4/8 |
| Ellipse | Ixx= πab³/4 Iyy= πa³b/4 Jo= πab/4(a²+b²) |
Created by:
shraddha9
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