Question

A 7300N elevator is to be given an acceleration of 0.150g by connecting it to a cable of negligible weight wrapped around a turning cylindrical shaft.
If the shaft's diameter can be no larger than 12.0cm due to space limitations, what must be its minimum angular acceleration to provide the required acceleration of the elevator

Answer 1

Answer:

The value is  $\alpha = 24.5 \ rad/s^2$

Explanation:

From the question we are told that

  The weight of the elevator is $W = 7300 \ N$

  The acceleration it is to be given is  $a = 0.150 \ g = 0.150 * 9.8 = 1.47 \ m/s^2$

  The diameter of the shaft is $d = 12.0 \ cm = 0.12 \ m$

Generally the radius is mathematically represented as

      $r = \frac{d}{2}$

=>   $r = \frac{0.12}{2}$

=>   $r = 0.06 \ m$

Generally the minimum angular acceleration is mathematically represented as

       $\alpha = \frac{a}{r}$

=>    $\alpha = \frac{1.47}{0.06}$

=>    $\alpha = 24.5 \ rad/s^2$