Question

A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. 11-48).A toy train of mass m is placed on the track and, with the system initially at rest, the train’s electrical power is turned on.The train reaches speed 0.15 m/s with respect to the track.What is the wheel’s angular speed if its mass is 1.1m and its radius is 0.43 m? (Treat it as a hoop, and neglect the mass of the spokes and hub.)

Answer 1

Answer:

0.166 rad/s

Explanation:

See attachment for calculations

Answer 2

Answer:

The angular velocity of the wheel is  $w = 0.1661 \ rad/ sec$

Explanation:

From the question we are told that

     The mass of the toy train is  $m$

     The speed of the train is $v_t = 0.15 m/s$

     The radius of the  wheel is  $r = 0.43 \ m$

     The mass of the wheel is $m_w = 1.1 * m$

According to the law of conservation of momentum

   $L_i = L_f$

Where $L _i$  is the initial angular momentum which is mathematically represented as

                 $L_i = rmv$

and

$L_f$ is the final angular momentum  which is mathematically represented as

                       $L_f = I * w$

Where I is the moment of inertia of the wheel which is mathematically represented  as

                 $I = m_w * r^2$

   So  

        $rmv = m_w r^2 w$

      $r * m * 0.15 = 1.1 * m * r^2 * w$

      $v = 1.1 * r * w$

But we know  the train is moving relative to the wheel so

     $v = v_t - wr$

Where wr is the linear velocity component of the wheel so

 Substituting values

        $0.15 - (w * 0.43) = 1.1 * 0.43 * w$

=>     $0.15 - (w * 0.43) = 0.473 * w$

        $0.15 = 0.903w$

       $w = 0.1661 \ rad/ sec$