Question

A rod of mass M = 3.5 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 45 g, moving with speed v = 5.02 m/s, strikes the rod at angle θ = 42° from the normal at a distance D = 2/3 L, where L = 1.2 m, from the point of rotation and sticks to the rod after the collision.

What is the initial angular momentum of the ball, in kilogram meters squared per second, right before the collision relative to the pivot point of the rod?

Answer 1

Complete Question

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Answer:

The initial angular momentum is  $L_i= 0.134 Kg .m^2/s$

Explanation:

From the question we are told that

        The mass of the rod is $M = 3.5 kg$

         The mass of the ball is $m =45g = \frac{45}{1000} = 0.045kg$

          The speed is $v = 5.02 m/s$

          The angle the ball strikes the rod is $\theta = 42^o$

           The distance from the center of the rod is $D = \frac{2}{3} L$

          The length L is  $= 1.2m$

          The initial angular momentum of the ball is mathematically represented as

                  $L_i = m\ v\ D\ cos \theta$

Substituting the value

                 $L_i = 45*10^{-3} * 5.02 * \frac{2}{3} * 1.2 * cos (42)$

                      $= 0.134 Kg .m^2/s$

     

Answer 2

Following are the calculation of the initial angular momentum of the ball:

Given:

$M = 3.5 \ kg \\\\m = 45\ g = 0.045\ kg\\\\ v = 5.02\ \frac{m}{s}\\\\ \theta = 42^{\circ}\\\\D = \frac{2}{3}\ L\\\\ L = 1.2\ m$

To find:

Li=?

Solution:

Using formula:

$Li= m v D \cos \theta=mv\times \frac{2L}{3}\cos \theta$

      $=0.045 \times 5.02 \times \frac{2 \times 1.2}{3}\cos 42^{\circ}\\\\=0.2259 \times\frac{2 \times 1.2}{3}\cos 42^{\circ}\\\\=0.18072\times \cos 42^{\circ}\\\\=0.13430113286\\\\=13.43 \times 10^{-2} \ \frac{kg\ m^2}{s}$

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