Torque acting on a small plane during a looping

In summary, the propeller needs to provide a torque to keep the plane in its vertical path. This torque is due to the angular motion, the loop, which is vertical to the loop.
  • #1
fara0815
45
0
Hello!
This physics problem seems unsolvable to me. I have been trying to figure it out myself but without any success. Maybe I am missing something obvious or I misunderstand the problem. I would be very happy to have any help:

The motorshaft and the popeller of a single engine plane have together a moment of inertia of 4.9 kg*m^2. While flying a loop with a radius of 180 m, the propellor has a frequency of 2100 min^-1 and the whole plane has a speed of 210 km/h, the plane needs to use its side rudder to produce a tourque in order to stay in its vertical path.
a) To which side does the pilot need to countersteer, if the propellor's angular velocity vector is pointed in the direction the plane is moving?
b) How great is the torque needed to keep it on its path (i.e. so the plane doesn't drift off to the side)?


It is not clear to me how to get a torque out of this picture. I think there is torque due to the angular motion, the loop, which is vertical to the loop. This torque could be responsible for the drift of the plane ?!
 
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  • #2
I have not figured the whole thing out, but I think you need to look at the change in angular momentum of the motorshart/propeller. Given the plane's speed and radius of the loop, you can figure out the rate of change of that angular momentum, and that will require a torque that the plane must provide. There will be a counter torque on the plane that will cause it to rotate if there is no control adjustment.
 
  • #3
thank you for your fast reply and your tips!
I am trying to figure it out the way you suggested. Wish me luck!
 
  • #4
I have not been able to solve it. I just cannot see why the angular moment of the propeller should change since it has nothing to do with the circular motion (the loop). The only thing that changes, at least as far as I get it, is the plane's horizontal distance to the center of the loop. But I have difficulties to make a connection between the distance and the propeller
 
  • #5
fara0815 said:
I have not been able to solve it. I just cannot see why the angular moment of the propeller should change since it has nothing to do with the circular motion (the loop). The only thing that changes, at least as far as I get it, is the plane's horizontal distance to the center of the loop. But I have difficulties to make a connection between the distance and the propeller
Angular momentum is a vector. Assuming the propeller is rotating clockwise as viewed from the cockpit, it points in the direction the plane is moving. If you change the direction of the plane by looping it, you are changing the direction of the angular momentum vector, and a torque is required to do that. The torque is the rate of change of angular momentum. In half a loop, the direction changes by 180°, so how much is the change in angular momentum? Don't forget that the change in a vector is also a vector.
 
  • #6
Thank you very much for your help!
Because of you I could finally solve it! I was thinking too complicated I guess.
Here is how I did it:

[tex]\omega_p =220 s^{-1}[/tex]
[tex]\omega_L =58.3 s^{-1}[/tex]

The angular velocity of the loop is constant, the propeller's isn't. It depends on the plane's location within the looping, the center of the looping is also the center of my coordinate system.

[tex]\omega_p=|\omega_p |\left(\begin{array}{cc}-sin(\omega_l t)\\cos(\omega_l t)\end{array}\right)[/tex]

The propeller's angular momentum is:
[tex]L_p=J\omega_p =J\omega_p =|\omega_p |\left(\begin{array}{cc}-sin(\omega_l t)\\cos(\omega_l t)\end{array}\right)[/tex]

The torque is the derivative with respect to the time of [tex]L_p[/tex]
[tex]M_p=J|\omega_p ||\omega_l | \left(\begin{array}{cc}-cos(\omega_l t)\\-sin(\omega_l t)\end{array}\right) [/tex]

[tex]\left(\begin{array}{cc}-cos(\omega_l t)\\-sin(\omega_l t)\end{array}\right)[/tex] is a unit vector and equal to 1.

So the equation for the torque becomes very simple:

[tex]M_p=J|\omega_p ||\omega_l = 349 Nm[/tex]

That matches the answer given by my professor!

I learned a lot with this one!
 
Last edited:

1. What is torque?

Torque is a measure of the force that causes rotational motion. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation.

2. How does torque affect a small plane during a looping?

During a looping, the torque acting on a small plane is constantly changing as the direction of the force and the distance from the axis of rotation change. This can cause the plane to rotate and experience changes in its velocity and direction of motion.

3. What factors can affect the torque acting on a small plane during a looping?

The force applied, the distance from the axis of rotation, and the angle between the force and the lever arm are all factors that can affect the torque acting on a small plane during a looping. Additionally, the shape and weight distribution of the plane can also play a role.

4. How can the pilot control the torque on a small plane during a looping?

The pilot can control the torque acting on a small plane during a looping by adjusting the force applied and the angle of the plane's wings. This can be done through the use of control surfaces, such as ailerons and elevators.

5. What are the potential risks of torque during a looping for a small plane?

If the torque on a small plane during a looping becomes too great, it can cause the plane to lose control and potentially crash. Additionally, the repeated changes in velocity and direction can put strain on the plane's structure and potentially cause damage. Proper training and understanding of torque is important for safely executing a looping maneuver.

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