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sergiy2304 [10]
3 years ago
11

A rod of mass M and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m, moving w

ith speed V, strikes the rod at angle θ from the normal and sticks to the rod after the collision. show answer No Attempt 50% Part (a) What is the total moment of inertia, I, with respect to the hinge, of the rod-ball-system after the collision?
Physics
1 answer:
Nastasia [14]3 years ago
5 0

Answer:

  I_total = L² (m + M / 3)

Explanation:

The moment of inertia is defined by

        I = ∫ r² dm

It is appreciated that it is a scalar quantity, for which it is additive, in this case the system is formed by two bodies and the moment of inertia must be the sum of each moment of inertia with respect to the same axis of rotation.

The moment of inertia of a bar with respect to an axis that passes through ends is

            I_bar = 1/3 M L²

The moment of inertia of a particle is

           I_part = m x²

We have to assume the point where the particle sticks to the bar, suppose it sticks to the end

              x = L

Total moment of inertia is the sum of these two moments of inertia

          I_total = I_bar + I_particule

 

          I_total = 1/3 M L² + m L²

         I_total = L² (m + M / 3)

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