Question

4 (Around a merry-go-round) rontal platform in the shape of a circular disk rotates in a horirontal plane about a frictionless, vertical axis through the center of the disk. The platform has a mass M 100 kg and a radius R = 1.9 m. Two students, whose masses are each m 54 kg (treat them each as a point mass), are located on the disk at radius T = 1.6 m and 0.9 m as shown in the figure. A. What is the total moment of inertia of the system? Provide your answer in kg.m? y dish & I student Istuent I = 1805 + 43.74 +138,24 student mR? 54 y 1.6% = 138.24 I - 3602.18 Kg.mo I student=54% Student=5480.92.42 Idisky $700 1m2² - 1/2100 x 1.92 Idiska hoo ama 8. The two children now move to the center of the disk, what is the total moment of Trentia of the system? Provide your answer in kg.m. I = 180.5 +OHO I, - 180.5 kg. m² I student, = 5,4x² = 0 I stucent, = 54xoroo I dish = (Yaloo X1.9² = 180.5

Answer 1

Ansuur ginen, Mass of platform M = lookg Radius of blatform R = 1.9m Mass of student = 54 kg Ą mo) jemo 0.9m 1. The total momentum of invitia of the system (It) It = I platform + I student, + Istudenz = momentum of invitea of the dise about an . anis berpendicular to it and therough its = ME? centre. M= mass of dise R = Radius of dise I platfoum = MR? - 100x (1•9) = 180.5 Kg m2 2 Istudent i = MR? = 5m= mass of student L R = distance from anis of Rotation ] = 54x(1.6)2 = 138.24 Kg m2. I student z = MR² = 54x(0.9² = 0.81 x 54 = 43.74 m

IT = 180.5+ 43.74 +138.24 It = 362.48 kg.m2 students at the centre of platform llas * The two children are at the center of 1 dise * distance of student from centre of dist is o: I student, = 54x(0)? =0 I student2 = 54x (0)2 0 I disc = I (100) (1.9)? . = 180.5 Kg.me IT = = 180.5+0+0 180.5 Kg.m²
answer : for part A = 362.48 kg.m^2

For part B = 180.5 kg.m^2