Problem 6-14: The centripetal acceleration and a linear speed of a point on an edge of an ordinary workshop grindstone

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 cmr=7.50\ \text{cm}, and ω=6500 rev/min\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 cm=0.075 mr = 7.50 \ \text{cm} = 0.075 \ \text{m}
ω=6500 rev/min×2π rad1 rev×1 min60 sec=680.6784 rad/sec\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 aca_{c}. Basing on the given data, we are going to use the formula

ac=rω2a_{c} = r \omega ^{2}

Substituting the given values, we have

ac=rω2ac=(0.075 m)(680.6784 rad/sec)2ac=34749.2313 m/s2ac=3.47×104 m/s2  (Answer)\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.

ac=34749.2313 m/s2×g9.81 m/s2ac=3542.2254gac=3.54×103g  (Answer)\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, vv 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

v=rωv=r\omega

If we substitute the given values, we have

v=rωv=(0.075 m)(680.6784 rad/sec)  v=51.0509 m/sv=51.1 m/s  (Answer)\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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