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Grade: 12
        
Plz help me solve this question and also tell me how to proceed with reshaping and recasting question.. 
one month ago

Answers : (4)

Harsh
45 Points
							
Considering the first case:
As the disc is reshaped hence the moment of inertia of disc with the axis passins through centroidal(centre) and perpendicular to the plane will be 
I1=MR²/2
The disc is recasted into a ring with same axis position but radius 2r
Hence let
I2=M(2R)²
Ase we know radius of gyration is
I=MK²
Hence k=√I/M
Hence factor will be √I1/M÷√I2/M
=√MR²/2M÷√4MR²/M
=R/√2÷√2R=1/2√2
If we take I2/I1 then We get ans as 2√2
one month ago
Harsh
45 Points
							
Considering dering the first case:
As the disc is reshaped hence the moment of inertia of disc with the axis passins through centroidal(centre) and perpendicular to the plane will be 
I1=MR²/2
The disc is recasted into a ring with same axis position but radius 2r
Hence let
I2=M(2R)²
Ase we know radius of gyration is
I=MK²
Hence k=√I/M
Hence factor will be √I1/M÷√I2/M
=√MR²/2M÷√4MR²/M
=R/√2÷√2R=1/2√2
If we take I2/I1 then We get ans as 2√2
one month ago
Harsh
45 Points
							
 
 the moment of inertia of disc with the axis passins through centroidal(centre) and perpendicular to the plane will be 
I1=MR²/2
The disc is recasted into a ring with same axis position but radius 2r
 
I2=M(2R)². 
Ase we know radius of gyration is
I=MK²
Hence k=√I/M
Hence factor will be √I1/M÷√I2/M
=√MR²/2M÷√4MR²/M
=R/√2÷√2R=1/2√2
If we take I2/I1 then We get ans as 2√2
one month ago
Harsh
45 Points
							
 
As the disc is reshaped hence the moment of inertia of disc with the axis passins through centroidal(centre) and perpendicular to the plane will be 
I1=MR²/2
The disc is recasted into a ring with same axis position but radius 2r
let
I2=M(2R)²
Ase we know radius of gyration is
I=MK²
Hence k=√I/M
Hence factor will be √I1/M÷√I2/M
=√MR²/2M÷√4MR²/M
=R/√2÷√2R=1/2√2
If we take I2/I1 then We get ans as 2√2
one month ago
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