- #1
justagirl
- 29
- 0
Radius of Gyration? REALLY need help ASAP!
Hey -
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!
Hey -
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!