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
[tex]v = 25.14 \ m/s [/tex]
2
[tex]a = 4013.5 \ m/s^2[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 0.30 \ m[/tex]
The angular speed is [tex]w = 1600 \ rpm = \frac{ 1600 * 2 * \pi }{60 } = 167.6 \ rad /s[/tex]
Generally the is mathematically represented as
[tex]r = \frac{d}{2} = \frac{0.30}{2 } = 0.15 \ m[/tex]
Generally the linear speed is mathematically represented as
[tex]v = wr[/tex]
[tex]v = 167.6 * 0.15[/tex]
[tex]v = 25.14 \ m/s [/tex]
Generally the acceleration is mathematically represented as
[tex]a = \frac{v^2 }{r}[/tex]
[tex]a = \frac{25.14^2}{0.15}[/tex]
[tex]a = 4013.5 \ m/s^2[/tex]
