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Angelina_Jolie [31]
3 years ago
13

A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.3 m and moment of inertia

Physics
1 answer:
vivado [14]3 years ago
7 0

Answer:

(a) 0.426 rad/s

(b) Before 296.225 J and After 148.5 J

Explanation:

At the center of merry-go-round, the person's weight is 75 Kg with moment of inertia of 820 kgm2 and the initial angular velocity is 0.85 rad/s

At the edge, the new angular velocity gained depends on the new moment of inertia

Here, the final moment of inertia is given by Initial moment of inertia + MR^{2} where M is the mass and R is the radius of merry-go-round

Final moment of inertia=820+(75*3.3^{2})=1636.75

From the law of conservation of angular momentum as torque is zero

I_i\times\omega_1=I_f\times\omega_2  and making \omega_2 the subject then

\omega_2=\frac {I_i\times\omega}{I_f}

where I_i and I_f are initial and final moment of inertia, \omega_1 and \omega_2 are initial and final angular velocity.

Substituting the provided values

\omega_2=\frac {0.85\times 820}{1636.75}=0.425843898\approx 0.426 rad/s

(b)

Initial rotational kinetic energy is given by

0.5I_i\times \omega_1^{2}

KE_i=0.5\times 820\times 0.85^{2}=296.225 J

The final rotational kinetic energy is given by

0.5I_f\times \omega_2^{2}

KE_i=0.5\times 1636.75\times 0.426^{2}\approx 148.5 J

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