A child ( m c = 39 kg) is playing on a merry-go-round ( m m = 250 kg, R = 2.3 m)

Question

A child (mc = 39 kg) is playing on a merry-go-round (mm = 250 kg, R = 2.3 m) that is initially at rest. The child then jumps off in a direction tangent to the edge of the merry-go-round as shown in the figure below. The child has a speed of 5.0 m/s just before she lands on the ground.

(b) What is the initial angular momentum of your system?
kg · m2/s

(d) If the final angular velocity of the merry-go-round is f, what is the magnitude of the final angular momentum of the merry-go-round? Express your answer in terms of f. (Use the following as necessary: mm, f and R. Do not substitute numerical values; use variables only.)

Lf,m =

What is the magnitude of the final angular momentum of the child? Express your answer in terms of f. (Use the following as necessary: mc, f and R. Do not substitute numerical values; use variables only.)

Lf,c =

(e) What is the magnitude of the final angular velocity of the merry-go-round?
rad/s

Explanation / Answer

(b) angular momentum = Iw where I is moment of inertia and w is angular velocity

initial angular momentum of system is zero as merry go round is at rest

(c) by conservation of angular momentum

final angular momntum of system= initial angular momentum of system=0

angular momentum of merry go round = Iw

considering merry go round as a disc

I=mr^2/2

final angular momentum of disc=1/2(mm)(R^2)(wf) -> Answer for part (c)

(d)angular momentum of child = mc*v*R=448.5 kgm2/s

(e)so angular momenum of merry go round will be also 448.5 but in opposite direction

equalling 1/2(mm)(R^2)(wf) =448.5

wf=0.6782 radian/second

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