Answer:
[tex]3.48\cdot 10^4 m/s^2, 3542 g[/tex]
Explanation:
First of all, we need to convert the angular velocity into rad/s:
[tex]\omega = 6500 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=680.7 rad/s[/tex]
Now we can calculate the centripetal acceleration at the edge of the grindstone, given by
[tex]a=\omega^2 r[/tex]
where
r = 7.50 cm = 0.075 m is the radius
Substituting,
[tex]a=(680.7)^2(0.075)=3.48\cdot 10^4 m/s^2[/tex]
And since
[tex]g=9.81 m/s^2[/tex]
The centripetal acceleration expressed in multiples of g is
[tex]a=\frac{3.48\cdot 10^4}{9.81}=3542g[/tex]
