Since a person is not really a point mass, the moment of inertia calculated in the example is an approximation. To estimate the error, consider each person to be a solid sphere of radius R = 0.30 m with its center at the end of the board and determine an expression for the moment of inertia for this system. What is the percent error made in the moment of inertia by treating the two people as point masses? (Use 61 kg for the mass of the father, 25 kg for the mass of the daughter, 10 kg for the mass of the seesaw, and 7 m for the length of the seesaw.) You can follow the Example, but when you consider the people as spheres, you will need to replace the moments of inertia of the father and daughter. To determine I for each person, use the parallel-axis theorem, I = ICM + MDZ, where D is the distance from the center of mass of the object to the axis of rotation, in this case %. The moment of inertia 2 1 1 . ZMR . Calculate the moment of inertia both ways, and nd the percent error using M x 100%. ofa sphere isI = 5 Ispheres Click the hint button again to remove this hint. X Use the parallel axis theorem and the properties of the moment of inertia to calculate the percentage change in moment of inertia from modeling the non-pointlike nature of the people on the seesaw.% I Total = 1097. 436 kg.mz Sphere) Isphere = 1097- 436 Kg. m2 For point like behaviour Food = James = 1 X 10 X 7 2 Brod = 40. 84 Ky. M2 I'm , = m, ( 4 ) 2 ( when we Consider both person as point ) = 61 x ( 3 . 5 ) 2 I'm = 747. 25 Imz = m2 (zz ) 2 = 25 ( 3 . 5 )2 = 306. 25 Rtotal ( point ) = Food + I'm, + I'm2 = 40.84+ 747:25 + 306:25 Ipoint = 1094. 34 Isphere - Lpoint ) x 100 = 1097.436- 1094. 34 Iophate 1097. 436