# Is it theoretically possible for the orientation angle of a projectile to remain exactly equal to the orientation of velocity?

This question is sparked by my answer to this question: Is this simulation following real physics?

After examining the math, I don't see how it is theoretically possible for the situation simulated to happen.

The situation in question is whether or not a spear in projectile motion can stay oriented over the entire path such that the angle is equal to the angle of the velocity vector.

I start with proposing that the rate of change of the angle of the velocity vector must be a constant - there can be no torque applied during the projectile motion, only the initial angular velocity imparted at the release.

So taking a projectile, with say an initial velocity of <20, 20> - this would be at a 45 degree angle, the case where the condition intuitively seems most likely:

$\frac{dy}{dt} = -9.8t + 20$, $\frac{dx}{dt} = 20$

We find $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-9.8t + 20}{20}$

And an expression for the angle, then:

$\theta = \arctan{\frac{dy}{dx}}$

Graphing the derivative of this function over the time span of the projectile shows that it is NOT a constant. This implies that there would have to be some sort of torque being applied mid-air if the angular displacement is directly proportional to $\theta$. However, it IS a very shallow bowl.

First of all, is this analysis correct?

Second, if this implies that the condition is impossible, what happens when one throws a javelin or spear perfectly? Why is it so natural to do so?

## 1 Answer

If you assume that no torque is imparted to the object during flight then you are entirely correct about the fact that the spear direction can not match the velocity direction. It may cross, perhaps cross twice, and perhaps tangent, but it will be a very different behavior.

A front-loaded weight in significant atmosphere, however, has a strong tendency to orient itself along the direction of motion. Given a shape with some symmetry, if the center of mass is offset from the center point of the object, then air resistance will tend to point the CM to the front of the object. In the case of a stick, you can imagine a constant force per unit length from the air, meaning that the effective point of action is in the geometric middle. If the CM is not in that location, an effective torque will be present.

My favorite example of this is lawn darts. The CM is intentionally place very far in the front and the object has a very significant air resistance to mass ratio. It orients itself with the tail fin trailing behind no matter how you throw it.

• Yup, bow arrows do this, too. – Florin Andrei Jul 6 '11 at 19:26
• @Florin That would be an even better example. Obviously that would be the entire point of an arrow. – Alan Rominger Jul 6 '11 at 19:41
• To be exact moving the center of mass (CM) to the front is not necessary, as long as the center of pressure (CP) lies behind the CM and a fin or something exists that applies a restoring torque when the orientation is different from the velocity vector. – ja72 Jul 6 '11 at 21:02
• I didn't even think of arrows or lawn darts. The dart example makes sense. I was correct in assuming that the situation couldn't occur without torque, but there is a restoring torque that arises from air resistance. Thanks for the answer. – cemulate Jul 6 '11 at 22:58