Question

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

What is the acceleration of the point?

Answer 1

Step-by-step explanation:

The linear speed is given by:

`v=2\pi f r`

So, replacing the given values:

`v=2\pi(1600rpm)((0.30m)/(2))\\v=1507.96(m)/(min)*(1min)/(60s)=25.13(m)/(s)`

Since we have a circular motion, the acceleration is directed radially toward the centre of the circle, that is the centripletal acceleration, which is defined as:

`a_c=(v^2)/(r)\\a_c=((20.13(m)/(s))^2)/(0.15m)\\a_c=4210.11(m)/(s^2)`