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kingkong69
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Find the moment of inertia of a wire, AB, of mass M and length pi*a, which is bent into a semicircle, about AB.
Mr^2/b]
The mass of the wire is M=pi*a*m, where m is the mass per unit length of the rod. Then a small element, ds is regarded, of the circumference of the semicircle as being approxiamtely a particle of mass mds at a distance a(radius) from the axis. The moment of inertia of such a particle would be m*(a^2)ds. as ds tends to 0, we integrate 1*ds for Pi*a and 0, resulting m*pi*a^3, substituting M we get Ma^2.
The correct answer should be (Ma^2)/2
could anyone advise me my mistake? thanks in advance
Mr^2/b]
The mass of the wire is M=pi*a*m, where m is the mass per unit length of the rod. Then a small element, ds is regarded, of the circumference of the semicircle as being approxiamtely a particle of mass mds at a distance a(radius) from the axis. The moment of inertia of such a particle would be m*(a^2)ds. as ds tends to 0, we integrate 1*ds for Pi*a and 0, resulting m*pi*a^3, substituting M we get Ma^2.
The correct answer should be (Ma^2)/2
could anyone advise me my mistake? thanks in advance
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