Calculating Angular Speed of Playground Merry-Go-Round

AI Thread Summary
To calculate the new angular speed of the playground merry-go-round after a child hops on, apply the conservation of angular momentum principle. Initially, the system's angular momentum is determined by the moment of inertia (I = 250 kg.m^2) and the initial angular speed (10.0 rev/min). When the 25.0 kg child sits on the edge, the total moment of inertia increases to 350 kg.m^2. The new angular speed can be found by setting the initial angular momentum equal to the final angular momentum, allowing for the calculation of the new speed. Using these principles will yield the correct angular speed after the child joins the ride.
Husker70
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Homework Statement


A playground merry-go round of Radius R=2.00m has a moment of
inertia I = 250 kg.m^2 and is rotating at 10.0 rev/min about a frictionless
vertical axle. Facing the axle a 25.0kg 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?

Homework Equations


I = 250 kg.m^2
10.0 rev/min
25.0kg child
radius = 2.00m

I=r^2m
L = r x p
L = Iw

The Attempt at a Solution



I = r^2m
m = I/r^2
m = 250kg.m^2/(2.00m)^2 = 62.5kg

I added the 62.5kg + 25.0kg = 87.5kg

I = r^2m
= (2.00m)^2(87.5kg) = 350kg.m^2

I don't know what to do with this from here.
Thanks for any help,
Kevin
 
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Use the conservation of angular momentum
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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