Solution Found!
A uniform disk of radius R is cut in half so that the
Chapter 11, Problem 34E(choose chapter or problem)
A uniform disk of radius ?R? is cut in half so that the remaining half has mass ?M (Fig.a) (a) What is the moment of inertia of this half about an axis perpendicular to its plane through point ?A?? (b) Why did your answer in part (a) come out the same as if this were a complete disk of mass ?M?. (c) What would be the moment of inertia of a quarter disk of mass ?M? and radius ?R? about an axis perpendicular to its plane passing through point ?B? (Fig.b)? Figure: (a) Figure: (b)
Questions & Answers
QUESTION:
A uniform disk of radius ?R? is cut in half so that the remaining half has mass ?M (Fig.a) (a) What is the moment of inertia of this half about an axis perpendicular to its plane through point ?A?? (b) Why did your answer in part (a) come out the same as if this were a complete disk of mass ?M?. (c) What would be the moment of inertia of a quarter disk of mass ?M? and radius ?R? about an axis perpendicular to its plane passing through point ?B? (Fig.b)? Figure: (a) Figure: (b)
ANSWER:Solution 34E Step 1: a) The moment of inertia of a shape can be given by using the equation, 2 I = r dm In our problem, the limit of the integral is from 0 to R, where R is the radius of the semicircle dm - mass element which can be given by, dm = dA Where, - mass density dA - Small area dA = (2 r) dr Therefore, dm = 2 r dr R 2 Therefore, I = r 2 r dr 0 R 3 I = 2 rdr 0 R r dr = R /4 4 0 I = 2 R /4 = ½ ( R ) R 2 ( R ) = M (total mass of the semi disc) 2 Moment of inertia, I = ½ MR