# Torque acting on a small plane during a looping

1. Nov 19, 2006

### fara0815

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 ?!

2. Nov 19, 2006

### OlderDan

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. Nov 19, 2006

### fara0815

thank you for your fast reply and your tips!
I am trying to figure it out the way you suggested. Wish me luck!

4. Nov 19, 2006

### fara0815

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. Nov 19, 2006

### OlderDan

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. Nov 21, 2006

### fara0815

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:

$$\omega_p =220 s^{-1}$$
$$\omega_L =58.3 s^{-1}$$

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.

$$\omega_p=|\omega_p |\left(\begin{array}{cc}-sin(\omega_l t)\\cos(\omega_l t)\end{array}\right)$$

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

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

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

So the equation for the torque becomes very simple:

$$M_p=J|\omega_p ||\omega_l = 349 Nm$$

That matches the answer given by my professor!

I learned a lot with this one!

Last edited: Nov 21, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook