Question

f the CD rotates clockwise at 500 rpmrpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 ss with constant angular acceleration, what is ααalpha, the magnitude of the angular acceleration of the CD, as it spins to a stop?

Answer 1

Answer:

The magnitude of the angular acceleration is  $a = 20.14 rad/s^2$

Explanation:

From the question we are told that

   The angular speed of CD is  $w_{CD} = 500 rpm = \frac{500 rpm}{\frac{60 \ s }{1 \ min} } * \frac{2 \pi }{ 1 \ rev} = 52.37 rad/s$

    time taken to decelerate is $t_{CD} = 2.60\ s$

    The final angular speed is  $w_f= 0 \ rad/s$

The angular acceleration is mathematically represented as

         $a = \frac{w_f - w_{CD}}{t}$

substituting values

          $a = \frac{0 - 52.37}{2.60}$

         $a = - 20.14 rad/s^2$

The negative sign show that the CD is decelerating  but the magnitude is

       $a = 20.14 rad/s^2$