Question

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

Answer 1

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

$0.5I_f\times \omega_2^{2}$

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