To solve this problem it is necessary to apply the concepts related to the relationship between tangential velocity and centripetal velocity, as well as the kinematic equations of angular motion. By definition we know that the direction of centripetal acceleration is perpendicular to the direction of tangential velocity, therefore:
[tex]a_c= \frac{v^2}{r}=r\omega^2[/tex]
Where,
V = the linear speed
r = Radius
[tex]\omega =[/tex] Angular speed
The angular speed is given by
[tex]\omega =6500\frac{rev}{min}(\frac{2\pi rad}{1 rev})(\frac{1 min}{60s})[/tex]
[tex]\omega = 680.6784 rad/s.[/tex]
Replacing at our first equation we have that the centripetal acceleration would be
[tex]a_c = r\omega^2[/tex]
[tex]a_c = (0.0750)(680.6784)^2[/tex]
[tex]a_c = 34 749.23 m/s^2[/tex]
To transform it into multiples of the earth's gravity which is given as [tex]9.8m / s[/tex] the equivalent of 1g.
[tex]a_c =34749.23 \frac{m}{s^2} (1\frac{g}{9.8m/s^2})[/tex]
[tex]a_c = 3545.84g[/tex]
PART B) Now the linear speed would be subject to:
[tex]v = \omega r[/tex]
[tex]v= (680.6784)(0.0750)[/tex]
[tex]v=51.05 m/s[/tex]
Therefore the linear speed of a point on its edge is 51.05m/s
