Since a person is not really a point mass, the moment of inertia calculated in t
ID: 2992872 • Letter: S
Question
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 radiusR=0.40m 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? (Use70kg for the mass of the father,25kg for the mass of the daughter, 10 kg for the mass of the seesaw, and 7 m for the length of the seesaw.)
Explanation / Answer
If = 2/5mf * rf^2 = 2*70*0.4^2/5 = 4.48 kgm^2 Id = 2/5md * rd^2 = 2*25*0.4^2/5 = 1.6 kgm^2 I sys = 1/12*10*7^2 + 4.48 + 70*3.5^2 + 1.6 + 25*3.5^2 = 1210.66 kg m^2 taking them as point mass I = 1/12*10*7^2 + 70*3.5^2 + 25*3.5^2 = 1204.58 kgm^2 %err = (1210.66 - 1204.58)/1210.66 * 100 = 0.5 %