Question

A high-speed drill rotating ccw at 2300 rpm comes to a halt in 2.60 s.

A) What is the magnitude of the angular acceleration of the drill?

B) How many revolutions does it make as it stops?

Please do not skip any details because I am very confused, I will appriciate it.

Answer 1

Initial angular velocity =2300 rpm
                         =2300 revolutions per minute
                    w=2300 *2*phi/60 sec
                    w=241 rad /s
It will come to halt in 2.60 sec
so final velocity becomes zero.

A) angular accelareation ( )=(w-w_o)/t
                                =(0-241)/2.60
                                =-92.37 rad/s^2
    ******************************

B)w^2-W^2=2**

=(w^2-W^2)/(2*alpha)

=314.27 rad

\Theta=50.017624 revovution

****************************************( 2phi rad=rev)

=50 revoutions

The body will come to halt after 50 revolutions