Question

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?

Answer 1

As we know that angular speed is given as

`\omega_i = 36 rad/s`

`\omega_f = 34.2 rad/s`

times taken as

`\Delta t = 0.595 s`

now we have

`\alpha = (\omega_f - \omega_i)/(\Delta t)`

now we have

`\aplha = (34.2 - 36)/(0.595)`

`\alpha = -3.03 rad/s^2`

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

`N = ((\omega_f + \omega_i)t)/(4\pi)`

Now we have

`N = ((36 + 34.2)(0.595))/(4\pi)`

`N = 3.32 rev`