# Radius of Gyration? REALLY need help

1. Nov 17, 2004

### justagirl

Radius of Gyration? REALLY need help ASAP!!!

Hey -
:surprised
If any of you can help me with the following problem asap that would be awesome!!

A racquet consists of uniform lamina that occupies the region inside the right-hand loop of r^2 = cos 2theta on the end of a handle (assumed to be of neglible mass) corresponding to the interval -1<=x<=0. Find the radius of gyration of the racquet around the line x = -1. Where is its sweet spot?

I know the radius of gyration around the y axis can be found by the using the formula: radius of g = sqrt(Inertia around y axis / mass)

Is this the correct formula for mass?

m = int(0...pi/4)int (0...sqrt(cos2theta) r cos theta r dr dtheta?

If so, how would revolving it around the line x = -1 instead of x = 0 change it?

Thanks!

2. Nov 18, 2004

### Staff: Mentor

Rotating about x=0 would give you give different moment than rotating around x=-1 and different moment of inertia.

I think the limits of integration for theta are $$-\frac{\pi}{2}$$ and $$\frac{\pi}{2}$$.

IIRC, you want to find the center of mass of the racquet. The sweet spot is at the center of mass where the maximum transfer of momentum would occur.

Last edited: Nov 19, 2004
3. Nov 19, 2004

### Sirus

Also consider doing this using the formulas for an area rather than a mass:
$$R=\sqrt{\frac{I_{x}}{A}}$$
where $I_{x}$ is the moment of inertia of the area relative to the x axis.

Last edited: Nov 19, 2004
4. Nov 19, 2004

### Staff: Mentor

Correction - the limits of $$\theta$$ are $$-\frac{\pi}{4}$$ and $$\frac{\pi}{4}$$. Been awhile since I have done this stuff - I should have graphed it first.

The first step is to find the mass moment of inertia $$I_c$$ with respect to the area's center of mass (remembering that the handle is massless).

Then the moment of inertia through a parallel axis is just $$I = I_c + md^2$$, where $$m$$ is the mass of the area in question, and $$d$$ is the distance between the centroid (center of mass of the area) and axis about which the object would rotate.

So the handle goes from -1 to 0, and the lamina goes between 0 to 1. Let the distance from x=0 to the lamina centroid by $$d_c$$, then $$d\,=\,1\,+\,d_c$$

Then once you know $$I$$,

radius of gyration g = $$\sqrt{\frac{I}{m}}$$.

If the axis of rotation was the y-axis (x=0), then $$d\,=\,d_c$$, and the radius of gyration would be less.

Remember, the greater the moment arm the more resistance to rotation, i.e. it would take more torque to get the same angular velocity.

Last edited: Nov 19, 2004