Question

Suppose an adult bowling ball weighs 12 pounds and has some radius R. This bowling ball has a moment of inertia Io. You don't have to determine the value of this inertia. Suppose a child's bowling ball has half the radius of the adult one, and is made of the same material. What is the rotational inertia of this ball, expressed as a fraction of 10? Express this as a fractional amount of I., such as 0.045 or 0.124.

Answer 1

Moment of Inertia of a solid sphere=

$\frac{2}{5}MR^{2}$ where M= mass of sphere & R= radius of sphere.We know that Mass= Density * Volume.

(K) Data Given IP = 0.453592 pounds, Solid sphere: weight = 12 Radius = R. M.O.I 1o. 12 P = 5.44311 kg 12 P = 5.443 (M = 5.44311 kg " # Moment of Inertia 2 MR² of solid sphere is Mass - Density * Volume M = Maps in kg. R= Radius. Mo = 8x UAR Io - 2 MR? T + 2 + 5. 44311 + R2 Mass of Child Bowling Ball S 4 (2) S4 A R x 1 (mo) Moment of Inertia of 2 mor- Child Bowling Bowl * 2 Mo (3) 254931 *R? BARCEREARS a254431*R**(32) - TO I! 0.03125 I - The moment of Inertia of child Bowling bowl is (0.031251)

Volume= $\frac{4}{3}\prod R^{3}$

Here the density is same since the material is same. mass of child bowling bowl will be different because of different radius. The radius is half of intial and due to cube power it will be 1/8 times of original mass.

All other necessary equation and its solving part is uploaded in image section. Thank You.