Problem 34 - College Physics

Question

\(\cdot\) A model of a helicopter rotor has four blades, each 3.40 \(\mathrm{m}\) in length from the central shaft to the tip of the blade. The model is rotated in a wind tunnel at 550 rev/min. (a) What is the linear speed, in \(\mathrm{m} / \mathrm{s}\) , of the blade tip? (b) What is the radial acceleration of the blade tip, expressed as a multiple of the acceleration \(g\) due to gravity?

Step-by-Step Solution

Answer

(a) The linear speed is 196.6 m/s. (b) The radial acceleration is approximately 1152 times gravity.

Step 1: Convert Revolutions per Minute to Radians per Second

First, we need to convert the rotor speed from revolutions per minute (rev/min) to radians per second (rad/s). One revolution is equal to \(2\pi\) radians and one minute is 60 seconds. Thus, \(550 \text{ rev/min} = 550 \times \frac{2\pi}{60} \text{ rad/s}\). This calculates to approximately \(57.6 \text{ rad/s}\).

Step 2: Calculate Linear Speed of Blade Tip

The linear speed \(v\) of the tip of the blade is given by the product of the angular speed \(\omega\) and the radius \(r\). Thus, \(v = \omega r\). With \(r = 3.40 \text{ m}\) and \(\omega = 57.6 \text{ rad/s}\), the linear speed is \(v = 57.6 \times 3.40 \approx 196.6 \text{ m/s}\).

Step 3: Calculate Radial Acceleration

The radial (centripetal) acceleration \(a_r\) is calculated using \(a_r = \omega^2 r\). Substitute \(\omega = 57.6 \text{ rad/s}\) and \(r = 3.40 \text{ m}\) into the formula to get \(a_r = (57.6)^2 \times 3.40 \approx 11295 \text{ m/s}^2\).

Step 4: Express Radial Acceleration as a Multiple of Gravity

Finally, express the radial acceleration as a multiple of the acceleration due to gravity \(g \approx 9.81 \text{ m/s}^2\). Calculate \(\frac{a_r}{g} = \frac{11295}{9.81} \approx 1152.4\). Thus, the radial acceleration is approximately 1152 times the acceleration due to gravity.

Key Concepts

Angular Velocity

Angular velocity is a measure of how quickly an object rotates around a point or an axis. Think of it as how fast something is spinning. It's expressed in terms of radians per second (rad/s). One full revolution is equivalent to two pi ().
To convert from revolutions per minute (rev/min) to angular velocity in radians per second, we use the formula:

• Revolutions per minute (rev/min) to radians per second: \( ext{Angularg velocity} = ext{rev/min} imes \frac{2\pi}{60} \).

In our helicopter rotor problem, with the speed of 550 rev/min, the angular velocity is calculated as \( 550 imes \frac{2\pi}{60} \approx 57.6 ext{ rad/s} \).
Understanding this concept is crucial because angular velocity helps us determine how fast the tip of the blade moves, thereby connecting rotational motion to linear speed.

Linear Speed

Linear speed refers to how fast an object travels over a distance. When something is rotating, like the tip of a helicopter blade, linear speed describes how quickly it moves across a given radius.
The relationship between linear speed \( v \) and angular velocity \( \omega \) is described by the formula:

• \( v = \omega r \) - Where \( r \) is the radius (distance from the center of rotation to the point of interest).

In the given rotor problem, with a radius of 3.40 m, and angular velocity of 57.6 rad/s, we find the linear speed as \( v = 57.6 \times 3.40 \approx 196.6 \text{ m/s} \).
The concept is vital for assessing how fast parts of machinery, like blades, are moving, which is important for engineering safety and design.

Radial Acceleration

Radial acceleration, also known as centripetal acceleration, occurs when an object moves in a circular path, and it points towards the center of the circle. This acceleration keeps the object moving along its curved path, preventing it from flying off due to inertia.
The formula for radial acceleration \( a_r \) is:

• \( a_r = \omega^2 r \) - \( \omega \) is angular velocity, and \( r \) is the radius.

In the helicopter rotor scenario, substituting \( \omega = 57.6 ext{ rad/s} \) and \( r = 3.40 ext{ m} \) gives \( a_r = (57.6)^2 \times 3.40 \approx 11295 \text{ m/s}^2 \).
Understanding radial acceleration helps in designing secure systems, making sure the structure can handle the forces exerted during rotation.

Centripetal Force

Centripetal force is essential for any object moving in a circular path. It is the actual force required to keep an object following its curved trajectory. Without it, the object would tend to move off in a straight line.
Centripetal force \( F_c \) can be calculated from radial acceleration using the equation:

• \( F_c = m \times a_r \) - Where \( m \) is the mass of the object, and \( a_r \) is its radial acceleration.

Although not requested in the helicopter rotor problem, knowing that centripetal force is what our rotor blades need to sustain their circular path is important for applications involving safety and stability of rotating machinery.
This concept plays a key role in various scenarios, from amusement park rides to planetary orbits, ensuring that forces at play match the motion observed.