Question

Se dispara una bala con una velocidad de v = 300[m/s] contra un cascarón esférico de papel que gira con MCU respecto a un eje vertical. Sabiendo que el radio del cascarón es de 10[cm]. Calcular la rapidez angular mínima que deberá girar el cascarón para que el proyectil haga únicamente un agujero. La dirección del movimiento de la bala pasa por el centro de la esfera.

Answer 1

Answer:

 w = 4.712 10⁻³ rad / s

Explanation:

For this exercise, the time it takes for the bullet to travel the distance of 2R must be equal to the time that the hole must travel half a circle.

Let's start by calculating the time it takes for the bullet, which is going at constant speed.

         v = x / t

         t = x / v

         t = 2R / v

         t = 2 0.10 / 300

         t = 6.666 10⁻⁴ s

As they ask that a single hole is formed in this time, it must be rotated half a circle, that is, θ =π rad, for which we use the angular scientific relations, where the shell has constant angular velocity

          w = θ / t

          w = π / 6,666 10⁻⁴

          w = 4.712 10⁻³ rad / s