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JEE Main 2021 (26 Aug, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2021 (26 Aug, Shift 1)

Q140JEE Main 2021 · 26 AUG · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

Consider a badminton racket with length scales as shown in the figure.

Figure

If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at r2\frac{r}{2} distance from the end A of the handle will be ................ Mr2.

Numerical answer type. Work it out, then open the answer.

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Answer: 52

  1. Handle: length 6r6r, centre 3r3r from A, so 2.5r2.5r from axis: M(6r)212+M(2.5r)2=9.25Mr2\frac{M(6r)^2}{12} + M(2.5r)^2 = 9.25Mr^2
  2. Ring: radius rr, centre 7r7r from A, so 6.5r6.5r from axis; axis is parallel to a diameter.
  3. Ring: Mr22+M(6.5r)2=42.75Mr2\frac{Mr^2}{2} + M(6.5r)^2 = 42.75Mr^2
  4. Total =52Mr2= 52Mr^2, so the answer is 5252.
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