Question

1. A person of mass 75.0 kg stands at the center of a rotating merry-go-round platform of radius 3.00 m and moment of inertia 826 kg⋅m2. The platform rotates without friction with angular velocity of 0.955 rad/s. The person walks radially to the edge of the platform. You may ignore the size of the person.

(a) Calculate the angular velocity when the person reaches the edge of the merry-go-round.

(b) Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.

Answer #1

we know the formula for angular momentum

P=I

so initial angular momentum

P_{0}=I_{0}_{0}....................1)

final angular momentum

P_{1}=I_{1}_{1}

I_{1}=I_{0}+Mr^{2}( when person reaches
the edge of the merry go round platform , he adds his momentum to
the total momentum)

P_{0}=P_{1}, so

I_{0}_{0}=(I_{0}+Mr^{2})_{1}

_{1}=I_{0}_{0}/I_{0}+Mr^{2}=826kg.m^{2}0.955rad/s/(826kg.m^{2}+75kg*(3m)^{2})=0.526rad/s

a) answer is 0.526rad/s

b) we know the formula for kinetic energy

K.E=1/2I^{2}

so K.E before the persons
walk=1/2*826kg.m^{2}*0.955rad/s=394 J

and K.E after the persons
walk=1/2(826kg.m^{2}+75kg*(3m)^{2}*(0.526rad/s)=395J

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