Question

Three children are riding on the edge of a merry-go-round that is 162 kg, has a 1.60 m radius, and is spinning at 17.3 rpm. The children have masses of 17.4, 29.0, and 32.8 kg. If the child who has a mass of 32.8 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm? rpm

Answer 1

40;= 1.8124 radls Mod Merry-go-round Im = 2 mm ga mass of the merry-go- round mm = 162 kg Radius of the merry-go-round 8=1.6m Emibial angulor speed w; = 17.3 8pm= 17.2. xa i rad/s The masses of the childrem ma = 17:4kg Mc2 - 29 kg, me3=32.8 kg of the Im = à x162 x(1-6)= 812.56 Im = 207.36 kg.me Mos of childi fci = mx 8²- 17.4x(16) = 44-544 kg m2 mod of childa Ic2= mc28 29 x0.6)² = 74.24 kg_m2 MOI of child 3 Tezz mego?= 32-8XC1•6)= 83.968 kg.me Smitial total mod of system I. - Im + I + ezt fez Li - 207.36+44.544+ 7424+83.968 410.112 kg-m Angular momentums L1 = 9; 60; = 410.112 x 1.8124 Li= 743.287 kg-m/ After child moves (32.8 kg) to the centre Ic3=0 kg .m2 final mol of the system If Imtfetica [Ef= 326.144 kg_m2 conservation of angular momentum Le = 24 743287 = 326.144xlof I, m? from

Wf=2-279 radls = 2.279 x60 rpm ат Ne 2 New angular speed wf = 21.75 rpm Initial angular speed wi-o radls (rest) Final angular speed wf = 1,00,000 rpm wf = 1,00,00° x an oad/s = 10,476.19 gadis Time t=2.5 min = 150 sec 150 b wf = wi pitat 10,476-19 = 0 + x 150) Q: 10.476.19 = 69.841 rad/s2 Angular acceleration d= 69.841 rad/s2 point at Y=8.5cm from the centre. x102m= Tangential acelenation a=8d=0.085x69-841 Frangential accelenation accelenation a = 5.94 mis Radial acceleration ay wogo ay=(10476.19) x(0.085) = 9328797.338 m/s 8= 8.5 X10 m= 0.085 m 2 uf ४ Radial acceleration ay = 93,28,797.338 mis ay = 93,28, 797,338 9.8 xg Radial acceleration ay=9,51,918. 19