# A cylinder and sphere rolling without slipping

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 [Broken]

Am I doing this right?

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

Doc Al
Mentor
NateTG said:
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 $\sin\theta$.