An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min.
(a) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of g.
(b) What is the linear speed of a point on its edge?
Solution:
We are given the following values: r=7.50\ \text{cm}, and \omega = 6500\ \text{rev/min} . We need to convert these values into appropriate units so that we can come up with sensical units when we solve for the centripetal acceleration.
r = 7.50 \ \text{cm} = 0.075 \ \text{m}
\omega = 6500 \ \text{rev/min} \times\frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1 \ \text{min}}{60\ \text{sec}} = 680.6784 \ \text{rad/sec}
Part A
We are asked to solve for the centripetal acceleration a_{c}. Basing on the given data, we are going to use the formula
Substituting the given values, we have
\begin{align*}
a_{c} & = r \omega ^2 \\ \\
a_{c} & = \left( 0.075 \ \text{m} \right) \left( 680.6784 \ \text{rad/sec} \right)^2 \\ \\
a_{c} & = 34749.2313 \ \text{m/s}^2 \\ \\
a_{c} & = 3.47 \times 10^{4} \ \text{m/s} ^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Now, we can convert the centripetal acceleration in multiples of g.
\begin{align*}
a_{c} & = 34749.2313 \ \text{m/s}^2 \times \frac{g}{9.81 \ \text{m/s}^2}\\ \\
a_{c} & =3542.2254g \\ \\
a_{c} & = 3.54\times 10^3 g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Part B
We are then asked for the linear speed, v of the point on the edge. So, we can use the given values to find the linear speed. We are going to use the formula
If we substitute the given values, we have
\begin{align*}
v & = r \omega \\ \\
v & = \left( 0.075 \ \text{m} \right)\left( 680.6784\ \text{rad/sec} \right) \ \ \\ \\
v & = 51.0509 \ \text{m/s} \\ \\
v & = 51.1 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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