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 mr=4.00\ \text{m}, and ω=300 rev/min\omega = 300 \ \text{rev/min}.

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

ω=300 revmin×2π rad1 rev×1 min60 sec=31.4159 rad/sec\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

ac=rω2a_{c} = r \omega ^2

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

ac=rω2ac=(4.00 m)(31.4159 rad/sec)2ac=3947.8351 m/s2ac=3.95×103 m/s2  (Answer)\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=rωv=r \omega

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

v=rωv=(4.00 m)(31.4159 rad/sec)v=125.6636 m/sv=126 m/s  (Answer)\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.

125.6636 m/s340 m/s×100%=36.9599%=37.0%\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 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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