Question

A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev

Answer 1

Answer:

a

    $\alpha = 2327.7 \ rev/s^2$

b

   $\theta = 9124.5 \ rev$

Explanation:

From the question we are told that

    The maximum  angular   speed is  $w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s$

     The  time  taken is  $t = 2.8 \ s$

     The  minimum angular speed is  $w_{min}= 0 \ rad/s$ this is because it started from rest

     

Apply the first equation of motion to solve for acceleration we have that

       $w_{max} = w_{mini} + \alpha * t$

=>     $\alpha = \frac{ w_{max}}{t}$

substituting values

       $\alpha = \frac{40950.73}{2.8}$

       $\alpha = 14625 .3 \ rad/s^2$

converting to $rev/s^2$

  We have

           $\alpha = 14625 .3 * 0.159155 \ rev/s^2$

           $\alpha = 2327.7 \ rev/s^2$

According to the first equation of motion the angular displacement is  mathematically represented as

       $\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2$

substituting values

      $\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2$

      $\theta = 57331.2 \ radian$

converting to revolutions  

        $revolution = 57331.2 * 0.159155$

        $\theta = 9124.5 \ rev$