Question

Luc, who is 1.80 m tall and weighs 950 N, is standing at the center of a playground merry-go-round with his arms extended,

Answer 1

Answer:

The angular velocity is $w_f = 35.6\ rpm$

Explanation:

From the question we are told that

    The height of Luc is $h = 1.80 \ m$

     The weight of Luc is  $W = 950 \ N$

      The mass of the dumbbell is  $m_b = 4.0 \ kg$

     The diameter of the merry-go-round is $d_1 = 4 m$

      The diameter of the merry-go-round is $r_1 = \frac{4}{2} = 2 m$

      The distance of the hand from the center is $L = 85cm = \frac{85}{100} = 0.85\ m$

       The diameter of Luc is $d_2 = 40 c m = \frac{40}{100} = 0.4 m$

       The diameter of the merry-go-round is $r_1 = \frac{0.4}{2} =0. 2 m$

       The weight of the merry-go round is  $W_m = 1500 \ N$

      The steady speed of the merry go round is  $w_i = 35 rpm = 35 * \frac{2 \pi }{60} = 3.66 rad/s$

Generally the moment of inertia before Luc brings his hand to his chest is mathematically represented as

            $I_i = \frac{m_b * r _1^2}{2 } + \frac{m_m * r^2_2}{y} + 2 * m_b L^2$

Where $m_m$ is the mass of the merry-go round which is

       $m_m = \frac{1500}{9.8} = 150 kg$

$m$ is the mass of Luc which is

       $m_m = \frac{950}{9.8} = 95 kg$

$m_b$ is the mass of dumbell which is

       $m_m = \frac{950}{9.8} = 95 kg$

So

    $I_i = \frac{150 * 2^2}{2} + \frac{95 * 0.2^2}{2} + 2 * 4 * (0.85)^2$

     $I_i = 307 .7 \ kg \cdot m^2$

The moment of inertia after Luc brings his hand to his chest is mathematically represented as

      $I _2 = \frac{m_b * r _1^2}{2 } + \frac{m_m * r^2_2}{y}$

Substituting value

       $I _2 = \frac{150 * 2^2 }{2 } + \frac{ 95 * (0.2) ^2}{2}$

       $I _2 = 302 \ kg \cdot m^2$

According to the law of conservation of angular momentum

      $w_f I_f = w_i I_i$

       $w_f = \frac{ w_i I_i}{I_f}$

 Substituting values

       $w_f = \frac{3.66 * 307.7}{302}$

       $w_f =3.73 \ rad/s$

Converting back to rpm

       $w_f =3.73 * \frac{60}{2 \pi}$

       $w_f = 35.6\ rpm$