- #1
SelHype
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A 3.0-m-diameter merry-go-round with rotational inertia 120 kg*m2 is spinning freely at 0.60 rev/s. Four 25-kg children sit suddenly on the edge of the merry-go-round. (a) Find the new angular speed in rev/s. (b) Determine the total energy lost to friction between the children and the merry-go-round in J.
Li[tex]\omega[/tex]i=Lf[tex]\omega[/tex]f
I=.5mR2
Okay so I know that Li[tex]\omega[/tex]i=Lf[tex]\omega[/tex]f. That means that (120 kg*m^2)(0.60 rev/s)=(mass of merry-go-round + 4(25kg))(1.52)[tex]\omega[/tex]f.
To find the mass of the merry-go-round I took the Inertia and divided it by R2 to get m = (120 kg*m2)/(1.52) = 53.3.
When I plugged it into the equation I found the new angular speed to be .21 rev/s.
My problem is with part (b)...
So I know that J is (kg*m2)/s2, but I am unsure as to how to solve for it...A little help, please?
Li[tex]\omega[/tex]i=Lf[tex]\omega[/tex]f
I=.5mR2
Okay so I know that Li[tex]\omega[/tex]i=Lf[tex]\omega[/tex]f. That means that (120 kg*m^2)(0.60 rev/s)=(mass of merry-go-round + 4(25kg))(1.52)[tex]\omega[/tex]f.
To find the mass of the merry-go-round I took the Inertia and divided it by R2 to get m = (120 kg*m2)/(1.52) = 53.3.
When I plugged it into the equation I found the new angular speed to be .21 rev/s.
My problem is with part (b)...
So I know that J is (kg*m2)/s2, but I am unsure as to how to solve for it...A little help, please?