# Radius of curvature of a Projectile

1. Aug 30, 2006

### konichiwa2x

Hi,
A particle is projeted with a velocity 'u' at an angle $$\theta$$ with the horizontal. Find the radius of curvature of the parabola traced out by the particle at the point where the velocity makes an angle $$\theta/2$$ with the horizontal.
I got $$u^2$$/gcos$$\theta/2$$ but my book gives a different answer.

2. Aug 30, 2006

### quasar987

Well, since you do not claim to have difficulty with the concept of radius of curvature or anything else, I sugest you post your work so we can see where you've gone wrong.

3. Aug 30, 2006

### konichiwa2x

radius of curvature = velocity^2/acceleration acting normal to the path towards the centre

when the angle the velocity makes with the horizontal becomes $$\theta/2 , v = ucos\theta/2 + u sin\theta/2$$

resul v = $$u^2$$
resolving 'g' along the perpendicular towards the centre, I got gcos$$\theta/2$$

so $$Radius of c = \frac{u^2}{gcos\theta/2}$$

Last edited: Aug 30, 2006
4. Aug 31, 2006

### data1217

$$v_y ^2= u_y ^2+ 2a_ys$$
Let the upward direction as positive
For the y-components and
$$\theta$$
$$0 = u^2{ \sin^2{\theta}} -2gh$$
$$h = \frac {u^2 {\sin^2{\theta}}}{2g}$$

For the $$\frac {\theta}{2}$$
$$h = \frac {u^2 {\sin^2{\frac {\theta}{2}}}}{2g}$$

Last edited: Aug 31, 2006
5. Aug 31, 2006

### andrevdh

The velocity should include both the x- and the y-velocity components. The normal acceleration is fine since the velocity is tangential to the path of the projectile. The x-component of the velocity will be $$u\cos(\theta)$$, but the y-component changes, since the projectile is accelerating in the y direction. The problem here is that the angle that the projectile makes with the origin is not $$\theta /2$$.

Last edited: Sep 1, 2006
6. Aug 31, 2006

### konichiwa2x

so is the velocity vector $$ucos\theta + usin\theta/2$$ ??

7. Sep 1, 2006

### andrevdh

For the y-component of the velocity at the point where its angle with the horizontal is $$\theta / 2$$ you can use the fact that

$$\tan(\theta /2) = v_y/v_x$$

This comes for the fact that the velocity vector at any stage in the motion is the resultant of the x-velocity component and y-velocity component of the motion. The angle theta/2 is the angle that the velocity vector is making at that stage with its x-velocity component.

To be mathematically correct you should rather write you equation in vector notation

$$\vec{v} = \hat{i}v_x + \hat{j}v_y$$

where the hat vectors are unit vectors in the direction of the x- and y-axes. Just writing a plus between the two components is misleading.

Last edited: Sep 1, 2006
8. Sep 1, 2006

### konichiwa2x

yup i did all that first itself

$$tan\theta/2 = \frac{usin\theta - gt}{ucos\theta}$$

so do i do this?
$$v = (ucos\theta) + (usin\theta - gt)$$

so do i take the magnitude of this and divide it by $$gcos\theta/2$$

Last edited: Sep 1, 2006
9. Sep 1, 2006

### andrevdh

Close, but you do not know the time. From the first equation in my previous post we have that

$$v_y = v_x \tan(\theta / 2)$$

which gives us the unknown y-velocity component in terms of the known x-velocity component.

Last edited: Sep 1, 2006
10. Sep 2, 2006

### konichiwa2x

ok thanks

so $$v = utan\theta/2j + ui$$

$$v^2 = u^2tan^2\theta/2 + u^2$$
$$v^2 = u^2sec^2\theta/2$$

$$radius of curvature = \frac{u^2sec^2\theta/2}{gcos\theta/2}$$

$$= \frac{u^2}{gcos^3\theta/2}$$

is it correct now?

11. Sep 2, 2006

### andrevdh

The x-component of the velocity stays the same during all of the motion of the projectile and is given by

$$u\cos(\theta)$$

which means that accoding to the formula in my presvious post that the y-component of the velocity at the required position will be

$$u\cos(\theta)\tan(\theta /2)$$

12. Jul 5, 2011

### kerimai

hi,
could you please help me,how to write an algorithm to calculate the radius is measured in steps of 3 degree of projected objects (irregular shaped objects in an image), then the algorithm will be converted to Matlab code.

13. Jul 5, 2011

### SteamKing

Staff Emeritus