A ballet dancer performs a spin by starting off in the position shown. The arms

Question

A ballet dancer performs a spin by starting off in the position shown. The arms can be represented as a single thin rod of mass ml=5kg and the length from hand to hand is D=1.2m (I = 1/12 MD^2), the horizontal leg can be represented by a horizontal rod mass m2=6kg and length R1 =1.(I = 1/3 MR_i^2), the rests of the body as a solid vertical cylinder of mass M=40kg and radius R=15cm. Write down an expression for the moment of inertia of the dancer's body in terms ml, m2, M, Rl,D, and R and calculate I. If the initial angular velocity is omega =3rad/s what is the angular momentum of the dancer?

Explanation / Answer

i)

moment of inertia = moment of inertia of hand + moment of inertia leg + moment of inertia body

I = (1/12)*m1*D^2 + (1/3)m2*Rl^2 + (1/2)*M*R^2

ii)

angular momentum L = I*w

L = ( (1/12)*m1*D^2 + (1/3)m2*Rl^2 + (1/2)*M*R^2)*w

L = (((1/12)*5*1.2^2)+ ((1/3*6*1^2) + ((1/2)*40*0.15^2))*3

L = 7.95 kg m^2 s^-1

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