Question

A helicopter rotor blade is \(7.80 \mathrm{~m}\) long and has a mass of 110 kg. (a) What force is exerted on the bolt attaching the blade to the rotor axle when the rotor is turning at 320 rev/min? (Hint: For this calculation the blade can be considered to be a point mass at the center of mass. Why?) (b) Calculate the torque that must be applied to the rotor to bring it to full speed from rest in \(6.70 \mathrm{~s}\). Ignore air resistance. (The blade cannot be considered to be a point mass for this calculation. Why not? Assume the distribution of a uniform rod.)

Short answer

The force on the bolt is 6800 N when the rotor is spinning at 320 rev/min and the torque needed to get the rotor to full speed from rest in 6.70 s is about 5400 N*m.

Step-by-step solution

Calculate the Centripetal Force

To calculate centripetal force, we first need to convert the rotational speed from revolutions per minute to radians per second. The formula to convert rpm to rad/s is \( \omega = \frac{2πN}{60} \) where \(N\) is the speed in rpm. After finding \( \omega \), we can calculate the centripetal force using the formula \( F = m \omega^2 r \) where \(m\) is the mass and \(r\) is the radius (or in this case, the length of the rotor blade).

Calculate the Moment of Inertia

Since in the second part we cannot consider the blade as a point mass, we have to calculate the moment of inertia 'I', which is the rotational equivalent of mass. For a rod rotating about one end, the moment of inertia is given by \( I = \frac{1}{3} m l^2 \) where \(m\) is the mass and \(l\) is the length of the rod.

Calculate the Angular Acceleration

We know the time taken to reach the final angular speed from rest, so we can calculate the angular acceleration \(\alpha\) using the formula \(\alpha = \frac{\omega}{t}\), where \(\omega\) is the final angular speed and \(t\) is the time taken.

Calculate the Torque

Now that we have both the moment of inertia and the angular acceleration, we can calculate the torque \(\tau\) using the formula \(\tau = I \alpha\)

Key concepts

Centripetal Force

Centripetal force is the required force to keep an object moving in a circular path, and it acts towards the center of the circle. In our helicopter rotor blade scenario, understanding centripetal force is crucial since it ensures the blade remains attached to the rotor axle while spinning. To compute the force on the bolt, we initially need to convert the rotor's rotational speed from revolutions per minute (rpm) to radians per second (rad/s), which is a common unit in physics for angular velocity.
Using the conversion formula \( \omega = \frac{2\pi N}{60} \), where \(N\) is the speed in rpm, gives us the angular velocity in rad/s. Once we have \(\omega\), we apply the formula \( F = m \omega^2 r \) to find the centripetal force, where \(m\) is the mass and \(r\) is the length of the rotor blade. This force is what the bolt experiences at a given rotational speed.

Moment of Inertia

The moment of inertia plays a pivotal role in rotational dynamics, analogous to mass in linear motion. It is a measure of an object's resistance to changes in its rotational motion. For the helicopter blade, we can't assume it to be a point mass since mass is distributed along its length, affecting how it rotates.
To calculate its moment of inertia (\(I\)), we use the formula \( I = \frac{1}{3} m l^2 \) for a rod that rotates around one end. Here, \(m\) is the mass and \(l\) is the length of the blade. The distribution of the blade’s mass along its length changes the dynamics compared to a point mass, which significantly impacts calculations in rotational motion. This property is crucial when determining the required torque to alter the blade's rotational speed.

Angular Acceleration

Angular acceleration (\(\alpha\)) is the rate at which angular velocity changes over time, equivalently as acceleration is to velocity in linear motion. In the context of the helicopter rotor achieving full speed, angular acceleration tells us how quickly the speed changes.
With the time (\(t\)) it takes to reach the final angular velocity (\(\omega\)), we determine angular acceleration using the formula \(\alpha = \frac{\omega}{t}\). Understanding and calculating angular acceleration is vital for assessing how fast the rotor blade speeds up or slows down, which impacts the performance and control of the helicopter.

Torque

Torque (\(\tau\)) is a measure of the force that causes an object to rotate about an axis and is a cornerstone of rotational dynamics. To find the amount of torque needed to bring our helicopter blade to full speed, we use the previously calculated moment of inertia and angular acceleration.
Applying the formula \(\tau = I \alpha\), we get the torque value. It’s essential to understand that torque is not merely about the force applied, but also how far from the rotation axis the force is exerted, which in the case of a blade is its length. This concept is crucial in designing and controlling systems with rotating parts such as helicopter rotors, engines, and wind turbines.

Rotational Speed Conversion

A common challenge in solving rotational dynamics problems is converting the rotational speed from one unit to another, such as from rpm to rad/s. This conversion is essential because it allows us to apply physics formulas that require angular velocity in rad/s.
With the helicopter blade rotating at 320 rpm, using the formula \( \omega = \frac{2\pi N}{60} \) enables us to convert that speed to rad/s, which is then used in formulas calculating centripetal force, torque, and other dynamics involved. Understanding how to accurately convert these measurements ensures precise calculations in determining the forces and dynamics of rotating systems.