Question

The wad of clay of mass m = 0.36 kg is initially moving with a horizontal velocity v1 = 6.0 m/s when it strikes and sticks to the initially stationary uniform slender bar of mass M = 4.3 kg and length L = 1.64 m. Determine the final angular velocity of the combined body and the x-component of the linear impulse applied to the body by the pivot O during the impact. The angular velocity is positive if counterclockwise, negative if clockwise. The impulse is positive if to the right, negative if to the left.

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The angular velocity is  $w_2 = 0.8479 \ rad /s$

The linear impulse applied to the body by the pivot O during the impact is

      $O_x \Delta t = 1.2153 N \cdot s$

Explanation:

The free body diagram of the image is shown on the second uploaded image

From the question we are told that

     The mass of the wad of clay is  $m = 0.36 \ kg$

      Th velocity is  $v_1 = 6.0 m/s$

      The mass of  slender bar $M = 4.3 \ kg$

        The length is  $L = 1.64 \ m$

From the diagram  the moment of inertia of the slender bar is

           $I_o = \frac{1}{12} ML^2 + M(\frac{L}{6} )^2$

And the moment of inertia of the  slender bar and the clay sticked to it is

       $I_T = \frac{1}{9} ML^2 + \frac{1}{9} mL^2$

            $= \frac{1}{9} [M+m]L^2$

According to the law of conservation of angular momentum

    The initial angular momentum =  final angular momentum

The initial angular momentum of the clay is  

             $P_i = mv_1 \frac{L}{3}$

The final angular momentum is

                   $P_f = \frac{1}{9}[M + m] L^2 w_2$  

 So

       $P_ i = P_f \equiv mv_1 \frac{L}{3} = \frac{1}{9} (M+ m )L^2 w_2$

So

        $w_2 = \frac{3mv_1}{(M+m)L}$

From the diagram the center of gravity is calculated as

       $r_G = \frac{M\frac{L}{6} + m \frac{L}{3} }{M+m}$

              $= \frac{M + 2m }{6 (M + m)} L$

Now angular velocity along the x-axis

      $-mv_1 + O_x \Delta t = -(M + m )r_G w_2$

     $-mv_1 + O_x \Delta t = -(M +m ) \frac{M + 2m}{6 (M+ m)} L w_2$

    $w_2 = \frac{3mv_1}{[M +m]L}$    

Where  $O_x \Delta t$  is  linear impulse applied to the body by the pivot O

Substituting values

       $w_2 = \frac{3 * 0.36 * 6}{[4.3 + 0.36] * 1.64}$

      $w_2 = 0.8479 \ rad /s$

   Making  $O_x \Delta t$  the subject of the formula

           $O_x \Delta t =\frac{M}{2( M + m )} mv_1$

substituting value

          $O_x \Delta t =\frac{4.3}{2( 4.2 + 0.36 )} (0.36)* (6.0)$

           $O_x \Delta t = 1.2153 N \cdot s$