A cylinder and sphere rolling without slipping

[scavok]

A hollow, thin-walled cylinder and a solid sphere start from rest and roll without slipping down an inclined plane of length 5 m. The cylinder arrives at the bottom of the plane 2.9 s after the sphere. Determine the angle between the inclined plane and the horizontal.

Ok, I think I'm close, but I'm getting really nasty equations at the end so I'm not very confident.

The acceleration of the sphere is 5*g*sin(x)/7
The acceleration of the cylinder is g*sin(x)/2

I'm pretty confident these are correct.

With the accelerations I should be able to use regular kinematics.

\Delta x=\frac{1}{2}at^2
Where \Delta x=5 and a=the acceleration of either the sphere or cylinder.

For the cylinder I get 5=\frac{1}{2}(\frac{gsin\Theta}{2})(t+2.9)^2
For the sphere I get 5=\frac{1}{2}(\frac{5gsin\Theta}{7})t^2

But when I solve for sin\Theta or t, and plug it into the opposite equation, I get nasty looking things like:
http://home.comcast.net/~andykovacs/equation.GIF

Am I doing this right?

 

[NateTG]

Since they roll (rather than sliding) you need to account for the energy that goes into making them spin.

 

[Doc Al]

Since they roll (rather than sliding) you need to account for the energy that goes into making them spin.

The equations already take that into consideration.

To scavok:

First solve for t by setting a_1 t^2 = a_2 (t + 2.9)^2 (the g \sin\theta will drop out).

Then use t to solve for \sin\theta.