Consider a badminton racket with length scale | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the handle be a rod of length $L = 6r$ and the ring be a circular hoop of radius $R = r$. The mass of both is $M$.$1$. Moment of inertia of th

Vedclass · Accepted Answer

$52$

Vedclass · Answer

(D) Let the handle be a rod of length $L = 6r$ and the ring be a circular hoop of radius $R = r$. The mass of both is $M$.$1$. Moment of inertia of the handle about an axis through its center of mass (perpendicular to it) is $I_{cm, rod} = \frac{M(6r)^2}{12} = 3Mr^2$.The distance of this axis from the rotation axis is $d_1 = (6r/2) - r/2 = 5r/2$.Using the parallel axis theorem,$I_{rod} = I_{cm, rod} + M(d_1)^2 = 3Mr^2 + M(5r/2)^2 = 3Mr^2 + 6.25Mr^2 = 9.25Mr^2$.$2$. Moment of inertia of the ring about its diameter is $I_{cm, ring} = \frac{MR^2}{2} = \frac{Mr^2}{2} = 0.5Mr^2$.The distance of the center of the ring from the rotation axis is $d_2 = 6r - r/2 + r = 6.5r = 13r/2$.Using the parallel axis theorem,$I_{ring} = I_{cm, ring} + M(d_2)^2 = 0.5Mr^2 + M(13r/2)^2 = 0.5Mr^2 + 42.25Mr^2 = 42.75Mr^2$.Total moment of inertia $I = I_{rod} + I_{ring} = 9.25Mr^2 + 42.75Mr^2 = 52Mr^2$.

Similar Questions

$A$ ring of mass $M$ and radius $R$ sliding with a velocity $v_0$ suddenly enters a rough surface where the coefficient of friction is $\mu$,as shown in the figure. Choose the correct statement$(s)$.

Four spheres,each of mass $M$ and diameter $2a$,are placed at the corners of a square of side $b$. The moment of inertia of this system about an axis along one of the sides of the square is:

$A$ meter stick is held vertically with one end on the floor and is allowed to fall. The speed of the other end when it hits the floor,assuming that the end at the floor does not slip,is ......... $m/s$ $(g = 9.8 \, m/s^2)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module