Question

1. What is the acceleration of a point on the edge of a 0.30 m diameter grinding wheel rotating at 1600 rpm?

Answer 1

Complete Question

1. What is the linear speed of a point on the edge of a 0.30 m diameter grinding wheel rotating at 1600 rpm?

2. What is the acceleration of the point?

Answer:

1

$v = 25.14 \ m/s$

2

$a = 4013.5 \ m/s^2$

Explanation:

From the question we are told that

    The  diameter is  $d = 0.30 \ m$

    The  angular speed is  $w = 1600 \ rpm = \frac{ 1600 * 2 * \pi }{60 } = 167.6 \ rad /s$

Generally the is mathematically represented as

         $r = \frac{d}{2} = \frac{0.30}{2 } = 0.15 \ m$

Generally the linear speed is mathematically represented as

     $v = wr$

     $v = 167.6 * 0.15$

      $v = 25.14 \ m/s$

Generally the acceleration is mathematically represented as

       $a = \frac{v^2 }{r}$

       $a = \frac{25.14^2}{0.15}$

         $a = 4013.5 \ m/s^2$