Hello,
I need to calculate the displacement of each end of a stick (or any cylindrical body) in which a force is exerted, as moment or torque somewhere between CG and one end, and rotates it around the center of gravity (or a fixed point). How many degrees does each side of the this lever move? What's the formula? Thank you in advance.
Mehdi

Hello Mehdi. The answer depends on several factors having to do with the material the stick is made from, it's cross-section shape and size, the location of the applied forces or moments, and the method of support. But first please clarify - are you applying a torque (twisting force) to this cylindrical stick, or do you mean that you are appying a normal force which causes bending of the stick? And how is the stick supported? Is it held at both ends, or just one end?

Thank you, ebaines, for you reply. Let me rephrase my question and mak it more tangible. The cylindrical body is not supported by anything. It's the tube of a flying model rocket which is exposed to wind, and the wind applies a normal force from one side and rotates it around the CG. We know the distance between CP (where the force is applied) and CG, mass of the rocket, the portion of mass of this section, and speed of the wind (and accordingly its mass and ultimately the force it exerts on the rocket). So I want to know how many degrees the rocket would rotate around the axis. Let's assume that wind is the only applied external force other than gravity. Last night, I realized that the formula is T = I*d_theta^2/dt^2 where T is torque, I is the moment of Inertia, d_theta is the increment of rotation angle, and dt is the time increment.
And T = F.d (N.m). F is the wind force and d is the ditance between CP and CG.
And I = m.r^2 where m is the portion of rocket's mass located between CP and CG.
So can you please tell me if I'm on the right track or I'm missing anything? I apprecite your help.

You are on the right track regarding torque and rotational acceleration


\Sigma T = I \alpha = I \frac {d^2 \theta}{dt^2}


But this equation assumes constant torque, under which the object doesn't just turn a few degrees but actually starts to spin in complete revolutions, going faster and faster as long as the torgue is non-zero. For the case of the rocket the force F caused by the side wind is reduced as the rocket rotates into the wind. Once the rocket is aligned with the prevailing wind it stops rotating because the torque would become negative if it did, causing it to rotate back to equilibrium. So the answer is - it depends on the wind speed and direction. Consider two cases:

1. An arrow in flight, which has no means of self propulsion, would rotate until it is aligned with the sum of the vectors representing its velocity across the ground plus the wind velocity. For example if it is shot at 50m/s and the wind is coming from the side at 25 m/s the arrow will turn to point 30 degrees from its flight path direction of travel.

2. A rocket that is self-propelled will veer offf course as it turns into the side wind, and change its tajectory. If there is no guidance mechanism the rocket will keep turning until it is aiming directly into the prevailing wind.

One last point - the value for I that you mentioned is valid only for a rotating disc. A cylinder or beam that rotates about its center has I = (1/12)mL^2, assuming the center of rotation is at the midpoint.

You are on the right track regarding torque and rotational acceleration


\Sigma T = I \alpha = I \frac {d^2 \theta}{dt^2}


But this equation assumes constant torque, under which the object doesn't just turn a few degrees but actually starts to spin in complete revolutions, going faster and faster as long as the torgue is non-zero. For the case of the rocket the force F caused by the side wind is reduced as the rocket rotates into the wind. Once the rocket is aligned with the prevailing wind it stops rotating because the torque would become negative if it did, causing it to rotate back to equilibrium. So the answer is - it depends on the wind speed and direction. Consider two cases:

1. An arrow in flight, which has no means of self propulsion, would rotate until it is aligned with the sum of the vectors representing its velocity across the ground plus the wind velocity. For example if it is shot at 50m/s and the wind is coming from the side at 25 m/s the arrow will turn to point 30 degrees from its flight path direction of travel.

2. A rocket that is self-propelled will veer offf course as it turns into the side wind, and change its tajectory. If there is no guidance mechanism the rocket will keep turning until it is aiming directly into the prevailing wind.

One last point - the value for I that you mentioned is valid only for a rotating disc. A cylinder or beam that rotates about its center has I = (1/12)mL^2, assuming the center of rotation is at the midpoint.

Thank you ebaines! Your answer was wonderful. You’re a rocket scientist!
My understanding is that, if the rocket is statically stable, meaning CP is below CG, the restoring force will return it back to stable position. Of course, as you mentioned, wind will change the trajectory and rocket will head against the direction of wind and peak altitude will be lower as expected. Please verify if I’m right.
In addition, I tried to find the equation for Moment of Inertia of this case, where the axis of rotation is not necessarily in the center and force is not exerted at the end but had no luck. If you can be kind enough to help me with that, I’d be very grateful.

I do believe that given a crosswind an unguided rocket will naturally turn into the wind, and hence its peak altitude will be less.

Moment of inertia has only to do wth geometry, not with where the force is applied. The general equation is


I = \int \rho r^2 dV


where \rho is the density of the material and r is the distance from the center of rotation. For a thin beam of homogenous materail rotating about its midpoint you get:


I = \int_{-L/2} ^{L/2} \rho A x^2 dx = \rho A\frac {L^3}{12} = \frac 1 {12} m L^2


If the axis of rotation is not in the center you can use the parallel axis theorem:


I_d = I_{cm} + md^2


where I_{cm} is the moment of inertia using the center of mass and d is the displacement of the center of rotation from the center of mass. For example if the center of rotation is the end of the beam you have d = L/2 which gives:


I_d = \frac 1 {12} mL^2 + m (L/2)^2 = \frac 1 3 mL^2

I do believe that given a crosswind an unguided rocket will naturally turn into the wind, and hence its peak altitude will be less.

Moment of inertia has only to do wth geometry, not with where the force is applied. The general equation is


I = \int \rho r^2 dV


where \rho is the density of the material and r is the distance from the center of rotation. For a thin beam of homogenous materail rotating about its midpoint you get:


I = \int_{-L/2} ^{L/2} \rho A x^2 dx = \rho A\frac {L^3}{12} = \frac 1 {12} m L^2


If the axis of rotation is not in the center you can use the parallel axis theorem:


I_d = I_{cm} + md^2


where I_{cm} is the moment of inertia using the center of mass and d is the displacement of the center of rotation from the center of mass. For example if the center of rotation is the end of the beam you have d = L/2 which gives:


I_d = \frac 1 {12} mL^2 + m (L/2)^2 = \frac 1 3 mL^2



Hi,
I hope I can still ask questions on this thread.
1. Considering the non-homogenous material of the rocket, do you think it’s fair to assume it as a thin beam as you explained, to simply the problem? If not, do I have to use the general equation for calculating Moment of Inertia? Is dV volume element?
2. In regards to the wind force, does it cause only a torque which exerts through CP and rotates the rocket around C.G. or it’s resolved in x-y-z axes too?If yes, in what coordinate system? Does it affect pitch angle as well? I have a hard time in projecting it onto the axes. Let’s say a rocket is flying with a pitch angle of 60 deg. With an azimuth angle of 50 deg measured from true north and a crosswind with an angle of 40 deg. (measured from true north) is hitting the rocket. Presumably the wind’s pitch angle is zero because its velocity vector is parallel to the local horizontal. The question is, how does it change the direction of rocket (azimuth) and pitch angle? I have the same question about drag and lift knowing that these two last forces do not change the direction of rocket (azimuth).
Thank you.

1. Yes, dV is the volume element, and hence if you have an equation for density as a function of position along the rocket's length you could calculate the moment of inertia accordingly.

2. The wind does two thing. First, it causes the rocket to change its course to veer into the wind - so if the wind is coming from the north horizontal to the ground that is the direction the rocket would ultimately tend to head in.. The rate at which the rocket turns is dependent on the torque caused by the cross wind acting on the tail fins. This force therefore depends on the relative speed and direction of the cross wind at each moment in time - i.e. as the rocket turns the torque is reduced. The second effect of the wind is to push the rocket down wind - the amount of deviation from its original course depends on the wind speed, cross-section of the rocket that faces the wind at each moment of time, and the drag coefficient of the rocket. The basic formula for drag is:

F_d = \frac 1 2 C_d A \rho v^2

where \rho = density of the air, C_d is the drag coefficient, and A is the cross-section facing the wind. In your case since the rocket is turning as it flies the cross-sectional area facing the wind varies over time and so does the coefficient of drag, so this is a very difficult computation. I would suggest that a numerical technique to determining F as a function of time is probably the best approach to evaluating this force, assuming you ghave data for C_d at all angles to the wind. Then you would add this force to the force of the rocket motor to determine the total force acting on the rocket at each moment, and determine the rockest's acceleration using \vec F(t) = m(t) \vec a(t). Keep in mind that the value for mass (m) varies with time as the rocket uses up is fuel, so you can't use the funadmental equations of motion that they teach in high school (i.e. d = \frac 1 2 a t^2 does not apply).

Bottom line is you have a very complicated set of circumsrtance that you are trying to model, and I'm wondering what the purpose of all this is? Are you working on a school project?

1. Yes, dV is the volume element, and hence if you have an equation for density as a function of position along the rocket's length you could calculate the moment of inertia accordingly.

2. The wind does two thing. First, it causes the rocket to change its course to veer into the wind - so if the wind is coming from the north horizontal to the ground that is the direction the rocket would ultimately tend to head in.. The rate at which the rocket turns is dependent on the torque caused by the cross wind acting on the tail fins. This force therefore depends on the relative speed and direction of the cross wind at each moment in time - i.e., as the rocket turns the torque is reduced. The second effect of the wind is to push the rocket down wind - the amount of deviation from its original course depends on the wind speed, cross-section of the rocket that faces the wind at each moment of time, and the drag coefficient of the rocket. The basic formula for drag is:

F_d = \frac 1 2 C_d A \rho v^2

where \rho = density of the air, C_d is the drag coefficient, and A is the cross-section facing the wind. In your case since the rocket is turning as it flies the cross-sectional area facing the wind varies over time and so does the coefficient of drag, so this is a very difficult computation. I would suggest that a numerical technique to determining F as a function of time is probably the best approach to evaluating this force, assuming you ghave data for C_d at all angles to the wind. Then you would add this force to the force of the rocket motor to determine the total force acting on the rocket at each moment, and determine the rockest's acceleration using \vec F(t) = m(t) \vec a(t). Keep in mind that the value for mass (m) varies with time as the rocket uses up is fuel, so you can't use the funadmental equations of motion that they teach in high school (i.e., d = \frac 1 2 a t^2 does not apply).

Bottom line is you have a very complicated set of circumsrtance that you are trying to model, and I'm wondering what the purpose of all this is? Are you working on a school project? .

askmehelpdesk.com doesn't notify me when I get a response.
Thank you very much for your all-inclusive reply. Yes this is for my master's thesis. I did my undergrad over 15 years ago so some of the concepts have escaped me.
Thanks anew.

askmehelpdesk.com doesn't notify me when I get a response.

Under "settings>edit options" check the susbription settings mode - should be set to the default of "instant email notification."

Thank you very much for your all-inclusive reply. Yes this is for my master's thesis. I did my undergrad over 15 years ago so some of the concepts have escaped me.
Thanks anew.[/QUOTE]

Yuo;e most welcome! Good luck with the thesis.

Under "settings>edit options" check the susbription settings mode - should be set to the default of "instant email notification."

Thank you very much for your all-inclusive reply. Yes this is for my master's thesis. I did my undergrad over 15 years ago so some of the concepts have escaped me.
Thanks anew.

Yuo;e most welcome! Good luck with the thesis.[/QUOTE]

Hi there again,
I have two questions:
In regards to torque and rotational acceleration equation:


\Sigma T = I \alpha = I \frac {d^2 \theta}{dt^2}


I need to derive {\theta} to find out how many degrees the body will turn by wind gust. I believe I need to integrate two times to obtain the rotational displacement (angle). Is this equation valid for this purpose, especially considering that I’m applying numerical integration using instant computations:
d_theta = wi*t + 1/2*a*t^2
where wi is the initial angular velocity.

Also, for moment of inertia, considering the mass of each section of the rocket (nose, recovery, motor fin etc.), I calculated their volume and density and accordingly the mass of infinitesimal elements of each section (assuming homogeneous material for each of them) and multiplied them by their distance from CG powered by 2 and added them up algebraically to obtain the total moment of inertia. Am I on the right track?
Thank you for your help.

I have two questions:
In regards to torque and rotational acceleration equation:


\Sigma T = I \alpha = I \frac {d^2 \theta}{dt^2}


I need to derive {\theta} to find out how many degrees the body will turn by wind gust. I believe I need to integrate two times to obtain the rotational displacement (angle). Is this equation valid for this purpose especially considering that I’m applying numerical integration using instant computations:
d_theta = wi*t + 1/2*a*t^2
where wi is the initial angular velocity.

Yes. And of course you also use \Delta w = at


Also, for moment of inertia, considering the mass of each section of the rocket (nose, recovery, motor fin etc.), I calculated their volume and density and accordingly the mass of infinitesimal elements of each section (assuming homogeneous material for each of them) and multiplied them by their distance from CG powered by 2 and added them up algebraically to obtain the total moment of inertia. Am I on the right track? Thank you for your help.

Yes - the moment of inertia can be calculated using


I = \sum _{i=1} ^N m_i r_i^2


where m_i is each bit of mass and r_i is the distance from the CG of that mass to the axis of rotation of the rocket.