Badminton is in shape of ring connected with a handle (a rod). Radius of ring is r, length of rod is 6r. If mass of ring is M, and of rod is m, find moment of inertia of system about the axis passing from a point at r/2 from end and perpendicular to ring plane.
(a) \(\frac{37}{4}mr^2\) + \(\frac{173}{4}MR^2\)
(b) \(\frac{37mr^2}{4} + Mr^2\)
(c) 3mr2 + 16Mr2
(d) 3mr2 + \(\frac{229}{4}Mr^2\)
Correct answer:(a) \(\frac{37}{4}mr^2\) + \(\frac{173}{4}MR^2\)
Explanation:
IO(ring) = Mr2 + M\(\Big(\frac{13}{2}r \Big)^2\)
= Mr2 + \(\frac{169}{4}Mr^2\)
= \(\frac{173}{4}Mr^2\)
Io(rod) = \(\frac{m(6r)^2}{12}mr^2\) + m\(\Big(\frac{5}{2}r \Big)^2\)
= 3mr2 + \(\frac{25}{4}mr^2\)
= \(\frac{37}{4}mr^2\)
Io(Badminton) = Io(ring) + Io(rod)
= \(\frac{37}{4}mr^2\) + \(\frac{173}{4}MR^2\)