Question

A 2.4 m long rod of mass m1 = 14 kg, is supported on a knife edge at its midpoint. A ball of clay of mass m2 = 3.5 kg is dropped from rest from a height of h = 1.4 m and makes a perfectly inelastic collision with the rod 0.9 m from the point of support. Find the angular momentum of the rod and clay system about the point of support immediately after the inelastic collision.

Answer 1

Explanation:

It is known that;

            $L_{i} = L_{f}$

Now, we need to calculate the value of L, that is, angular momentum.

Therefore,

                   L= mvr

where,   m = mass

              v = velocity

              r = radius

Hence,

               L = $m(\sqrt{2gh}) \times r$

So,

              L = $(2.2 \times (\sqrt{2 \times 9.81 \times 1.4}) \times 0.9)$

                  = 10.37 Js

Thus, we can conclude that angular momentum of the rod and clay system about the point of support immediately after the inelastic collision is 10.37 Js.