Question

From a uniform sphere of mass M and radius R a cavity of diameter R is created . find the ratio of moment of inertia of the sphere left about AA' and BB'

Answer 1

Answer:

Explanation:

In this question first time take a wholl area of circle and then cut section area is minus

That is A1x1-A2x2/A1-A2

Answer 2

The ratio of moment of inertia of sphere left about AA' and BB" is 62/57

• First we need to find the mass of the cavity . We will find the mass of the cavity by using unitary method
• It is given that the radius of the cavity is R/2
• 4πR³/3 unit volume → contains mass M
• 1 unit volume → contains mass M/(4πR³/3)
• ∴  4π(R/2)³/3 unit volume → contain mass [M/(4πR³/3)] × 4π(R/2)³/3
• ∴ 4π(R/2)³/3 unit volume contains mass M/8
• the moment of inertia along AA' axis = (MOI of the sphere along AA') - (MOI of the cavity along AA')
• the moment of inertia of a solid sphere along the diameter is (2/5)MR²
• MOI of the sphere along AA'=(2/5)MR²
• MOI of the cavity along AA'= (2/5)(M/8)(R/2)²=(1/160)MR²
• MOI along AA'= (2/5)MR² - (1/160)MR² = (31/80)MR²
• MOI along BB' = (MOI of the sphere along BB') - (MOI of the tcavity along BB')
• MOI of sphere along BB'=(2/5)MR²
• MOI of cavity along BB' =  (2/5)(M/8)(R/2)² + (M/8)(R/2)² =(7/160)MR²                               (using parallel axis theorem )
• MOI of along BB'= (2/5)MR² - (7/160)MR²   = (57/160)MR²
• Now the ratio of MOI about $\frac{AA'}{BB'}$=$\frac{(31/80)MR^2}{(57/160)MR^2}$      =   (62/57)