Merry go round Rotation Problem

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The discussion centers on calculating the initial angular velocity and angular acceleration of a merry-go-round that rotates 15 times in 45 seconds and is then stopped by a child in 15 seconds. The initial angular velocity is determined to be 1/3 of a revolution per second, which converts to 2/3 pi radians per second. Participants confirm the conversion from revolutions to radians and discuss the implications for further calculations. The conversation emphasizes the importance of understanding angular velocity in radians per second for solving the problem. Overall, the focus is on the correct application of angular motion formulas.
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Homework Statement


A merry - go - round rotates clockwise 15 times in 45 seconds. By rubbing against its edge, a child stop it from turning in 15 seconds.

a) find its initial angular velocity (ω)
b) find its angular acceleration (\alpha)


The Attempt at a Solution



15 times in 45 seconds means:

in 1 second it makes 1/3 of a revolution

thus in 3 seconds it make on complete revolution

im not sure if I am going in the right direction but Angular Velocity is in radians per second.

Am i on the right track? Do i just take the reciprocal of my answer? Please help
 
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You found that it makes 1/3 revolution per second. Can you convert 1/3 revolution to radians?
 
TSny said:
You found that it makes 1/3 revolution per second. Can you convert 1/3 revolution to radians?

2/3 pi radians?
 
Yes.
 
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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