A merry-go-round of radius r= 2.0 m and moment of 500 kg.m^2 rotates about a frictionless pivot. It makes one revolution every 5 seconds. A child of mass 25kg walks from the center (r=0) to the edge (r=R).
How long does it take the merry-go-round to make one revolution when the child is standing at the edge? (The moment of inertia of a circular disc is I=1/2mR^2)
Is this process elastic? (That is, is the kinetic energy the same when the child is at r=0 and r+R?) Explain your answer
A merry-go-round of radius r= 2.0 m and moment of 500 kg.m^2 rotates about a frictionless...
A playground merry-go-round of radius R = 2.10 m has a moment of inertia of I = 260 kgm2 and is rotating at 11.0 rev/min about a frictionless vertical axis. Facing the axle, a 23.0 kg child hops on to the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round?
A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 255 kg · m2 and is rotating at 8.0 rev/min about a frictionless vertical axle. Facing the axle, a 24.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round
A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 245 kg · m2 and is rotating at 12.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round? rev/min
A playground merry-go-round of radius R = 2.2C m has a moment of inertia I = 265 kg middot m^2 and is rotating at 8.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?
A disc-shaped merry-go-round has a mass of 100 kg and a radius of 1.60 m and is initially spinning at 20.0 rpm, with a 33-kg child sitting at its center. The child then walks out to the edge of the disc. (a) Find the initial and final moments of inertia of the system (disc + child), treating the child as a point particle. (b) State why the system’s angular momentum is conserved. (You can assume that the axis of the...
46. W A playground merry-go-round of radius R 2.00 m has a moment of inertia 1 250 kg m2 and is rotating at 10.0 rev/min about a frictionless, vertical axle. Facing the axle, a 25.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?
A playground merry-go-round of radius R = 2.20 m has a moment of inertia of I = 280 kgm2 and is rotating at 11.0 rev/min about a frictionless vertical axis. Facing the axle, a 21.0 kg child hops on to the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round? 8.04 rad/s 0.845 rad/s 3.17 rad/s 1.18 rad/s 0.445 rad/s
A playground merry-go-round has a radius of R = 4.0m and has a moment of inertia I_cm = 7.0 times 10^3 kg middot m^2 about an axis passing through the center of mass. There is negligible friction about its vertical axis. Two children each of mass m = 25kg are standing on opposite sides a distance r_0 = 3.0m from the central axis. The merry-go-round is initially at rest. A person on the ground applies a constant tangential force of...
#46 A playground merry-go-round of radius R=2.oom has a moment of Inertia I = 250kg, om² and is rotating at 10.0 rev/min about a frictionless, vertical axle. Facing the axle, a 250-kg Child hoops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?
need help ASAP please A playground merry-go-round of radius R = 2.10 m has a moment of inertia of I = 260 kgm2 and is rotating at 11.0 rev/min about a frictionless vertical axis. Facing the axle, a 23.0 kg child hops on to the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round? O 0.229 rad/s 1.21 rad/s 0.829 rad/s 0.829 rad/s 2.95 rad/s