Question

A playground merry-go-round of radius r = 1 m has a moment of inertia of i = 240 kg*m^2. and is rotating at a rate of ω = 8 rev/min around a frictionless vertical axis. facing the axle, a 35 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 (in rev/min)?

Answer 1

To find the new angular velocity of the merry-go-round after the child gets on, we can use the principle of conservation of angular momentum. By applying the conservation of angular momentum equation, we can calculate the new angular velocity of the merry-go-round as 0.345 rev/s.

Step-by-step explanation:

To find the new angular velocity of the merry-go-round after the child gets on, we can use the principle of conservation of angular momentum. Initially, the merry-go-round is rotating with an angular velocity of 0.500 rev/s and has a moment of inertia of 1000.0 kg*m². The child has a mass of 22.0 kg and is initially at rest. When the child grabs the outer edge of the merry-go-round, the system's moment of inertia increases. The total initial angular momentum of the system is given by L1 = Iω1, where I is the initial moment of inertia and ω1 is the initial angular velocity. The final angular momentum of the system with the child on it is given by L2 = (I + Ic)ω2, where Ic is the moment of inertia of the child and ω2 is the final angular velocity.

Applying conservation of angular momentum, we have L1 = L2, which gives us Iω1 = (I + Ic)ω2. Rearranging the equation and solving for ω2, we get:

ω2 = (Iω1) / (I + Ic)

Substituting the values, ω1 = 0.500 rev/s, I = 1000.0 kg*m², and Ic = MR² = (22.0 kg)(4.0 m)² = 352.0 kg*m², we can calculate the new angular velocity:

ω2 = (1000.0 kg*m² * 0.500 rev/s) / (1000.0 kg*m² + 352.0 kg*m²)

Answer 2

When the child jumps onto the merry-go-around the moment of inertia of the system changes. If we consider the child to be point-like mass then its moment of inertia would be:

`I_(ch)=mr^2`
We get the new moment of inertia by simply adding the child's moment of inertia to the old moment of inertia.

`I_(new)=I_(old)+I_(ch)=240+35(1)^2=275 $kgm^2`
Since there is no force mention we must assume that angular momentum is conserved.

`L=const.\\ L=I_(old)\omega_0=I_(new)\omega'\\ \omega'=(I_(old)\omega_0)/(I_(new))`
When we plug in all the numbers we get:

`\omega'=(240\cdot8)/(275)=6.98 \ (rev)/(min)`