- #1
Mirumbe
- 4
- 0
Find the moment of inertia of a uniform solid sphere of mass,m and radius,a about an axis through its centre.
I have tried to solve it but I get the different answer, I don't know where I have done mistake. Please! check and correct my solution below:-
Consider a volume element, dv of the sphere; this has mass,mdv/(4/3)∏a^3.
Then
I = ∫(mdv/(4/3)∏a^3)r^2
Where r is the distance of the volume element, dv from the axis, and
I is the moment of inertia
I = m/(4/3)∏a^3∫r^2dv
We know, the volume of the sphere,v is:
v = (4/3)∏r^3
dv = 4∏r^2dr
Thus,
I = 3m/(4/3)∏a^3∫r^2(4∏r^2)dr
= 3m/a^3∫r^4dr under limit [0,a]
= 3m/a^3[r^5/5] under [o,a]
Hence,
I = (3/5)ma^2 --------------WRONG!
The correct answer is: I = (2/5)mr^2
I have tried to solve it but I get the different answer, I don't know where I have done mistake. Please! check and correct my solution below:-
Consider a volume element, dv of the sphere; this has mass,mdv/(4/3)∏a^3.
Then
I = ∫(mdv/(4/3)∏a^3)r^2
Where r is the distance of the volume element, dv from the axis, and
I is the moment of inertia
I = m/(4/3)∏a^3∫r^2dv
We know, the volume of the sphere,v is:
v = (4/3)∏r^3
dv = 4∏r^2dr
Thus,
I = 3m/(4/3)∏a^3∫r^2(4∏r^2)dr
= 3m/a^3∫r^4dr under limit [0,a]
= 3m/a^3[r^5/5] under [o,a]
Hence,
I = (3/5)ma^2 --------------WRONG!
The correct answer is: I = (2/5)mr^2