Author Archives: Engr. Jonas Paul B. de la Cruz

Problem 6-15: The centripetal acceleration at the tip of a helicopter blade

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip.

(a) Calculate the magnitude of the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min.

(b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s).

Solution:

Part A

We are given the following values: r=4.00\ \text{m}, and \omega = 300 \ \text{rev/min}.

Let us convert the angular velocity to unit of radians per second.

\omega = 300 \  \frac{\text{rev}}{\text{min}} \times \frac{2\pi \ \text{rad}}{1 \ \text{rev}}\times \frac{1\ \text{min}}{60 \ \text{sec}} = 31.4159 \ \text{rad/sec}

The centripetal acceleration at the tip of the helicopter blade can be computed using the formula

a_{c} = r \omega ^2

If we substitute the given values into the formula, we have

\begin{align*}
a_{c} & = r \omega^2 \\ \\
a_{c} & = \left( 4.00\ \text{m} \right)\left( 31.4159 \ \text{rad/sec} \right)^2 \\ \\
a_{c} & = 3947.8351 \ \text{m/s}^2 \\ \\
a_{c} & = 3.95 \times10^3 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are asked to solve for the linear velocity of the blade’s tip. We are going to use the formula

v=r \omega

We just needed to substitute the given values into the formula.

\begin{align*}
v & = r \omega \\ \\
v & = \left( 4.00 \ \text{m} \right)\left( 31.4159 \ \text{rad/sec} \right) \\ \\
v & = 125.6636 \ \text{m/s} \\ \\
v & = 126 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Let us compare this with the speed of light which is 340 m/s. 

\frac{125.6636 \ \text{m/s}}{340\ \text{m/s}} \times 100 \%= 36.9599 \% =37.0\%

The linear velocity of the blades tip is 37.0% of the speed of light.

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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\ \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

a_{c} = r \omega ^{2}

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

v=r\omega

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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