tiannahines9240 tiannahines9240
  • 15-01-2020
  • Physics
contestada

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?

Respuesta :

sebasacevedop
sebasacevedop sebasacevedop
  • 15-01-2020

Explanation:

The linear speed is given by:

[tex]v=2\pi f r[/tex]

So, replacing the given values:

[tex]v=2\pi(1600rpm)(\frac{0.30m}{2})\\v=1507.96\frac{m}{min}*\frac{1min}{60s}=25.13\frac{m}{s}[/tex]

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:

[tex]a_c=\frac{v^2}{r}\\a_c=\frac{(20.13\frac{m}{s})^2}{0.15m}\\a_c=4210.11\frac{m}{s^2}[/tex]

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