meep26wesley
contestada

To throw a curve ball, a pitcher gives the ball an initial angular speed of 36.0rad/s. When the catcher gloves the ball 0.595s later, its angular speed has decreased (due to air resistance) to 34.2rad/s. What is the ball’s angular acceleration and how many revolutions does the ball make before being caught?

Respuesta :

As we know that angular speed is given as

[tex]\omega_i = 36 rad/s[/tex]

[tex]\omega_f = 34.2 rad/s[/tex]

times taken as

[tex]\Delta t = 0.595 s[/tex]

now we have

[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]

now we have

[tex]\aplha = \frac{34.2 - 36}{0.595}[/tex]

[tex]\alpha = -3.03 rad/s^2[/tex]

Now to find the number of revolutions we can use another equation

[tex]N = \frac{(\omega_f + \omega_i)t}{4\pi}[/tex]

Now we have

[tex]N = \frac{(36 + 34.2)(0.595)}{4\pi}[/tex]

[tex]N = 3.32 rev[/tex]

Otras preguntas