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vivado [14]
3 years ago
7

In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small spher

e of clay is dropped onto the disk, and the sphere sticks to the disk. All frictional forces are negligible. What would happened to the angular momentum and the total kinetic energy of the disk-sphere system immediately before and after the collision?
Physics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

  L₀ = L_f ,  K_f < K₀

Explanation:

For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.

This means that the angular momentum before and after the collision changes.

Initial instant. Before the crash

        L₀ = I₀ w₀

Final moment. Right after the crash

        L_f = (I₀ + mr²) w

we treat the clay sphere as a point particle

how the angular momentum is conserved

       L₀ = L_f

       I₀ w₀ = (I₀ + mr²) w

       w = \frac{I_o}{I_o + m r^2}   w₀

having the angular velocities we can calculate the kinetic energy

       

starting point. Before the crash

        K₀ = ½ I₀ w₀²

final point. After the crash

        K_f = ½ (I₀ + mr²) w²

sustitute

        K_f = ½ (I₀ + mr²)  ( \frac{I_o}{I_o + m r^2}   w₀)²

        Kf = ½  \frac{I_o^2}{ I_o + m r^2}   w₀²

we look for the relationship between the kinetic energy

        \frac{K_f}{K_o}=   \frac{I_o}{I_o + m r^2}

       \frac{K_f}{K_o } < 1

      K_f < K₀          

we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision

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