A cylinder and sphere rolling without slipping

1. Apr 6, 2006

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 [Broken]

Am I doing this right?

Last edited by a moderator: May 2, 2017
2. Apr 6, 2006

NateTG

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

3. Apr 6, 2006

Staff: Mentor

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 $\sin\theta$.