A merry-go-round with a a radius of R = 1.61 m and moment of inertia I = 215.0 k

Question

A merry-go-round with a a radius of R = 1.61 m and moment of inertia I = 215.0 kg-m2 is spinning with an initial angular speed of ? = 1.51 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 56.0 kg and velocity v = 4.6 m/s runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round. Calculated initial ang momentum of merry-go-round as 324.65 kg-m^2/s, initial ang momentum of person as 414.736, and final velocity of system as 2.053 rad/s 5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on? Please don't copy and paste the answer from Yahoo answers, it's incorrect.

Explanation / Answer

angular momentum, L = I

and for discrete masses= m(v X R)

initial angular momentum of the merry-go-round = I = 215*1.51 = 324.65 kg-m2/s

and initial angular momentum of the person= m(v X R)=mvR = 56*4.6*1.61 = 414.736 kg-m2/s

from the angular momentum conservation,

initial angular momentum= final angular momentum

initial angular moment = 324.65+414.736 = 739.386 kg-m2/s

and final moment of inertia,I = I+mR2 = 215+56*1.612 = 360.1576 kg-m2

let final angular velocity= final

therefore,

(739.386)=(360.1576)*final

hence, final = 2.053 rad/s

force need to person to hold on the merry-go-round= m2final*R= 56*2.0532*1.61 = 380 N

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