# A cuboid on an inclined plane - based on an Olympiad problem

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1. Mar 20, 2015

### marcnn

1. The problem statement, all variables and given/known data
Let's suppose we have a [cuboid](http://en.wikipedia.org/wiki/Cuboid) of dimensions $a \times b \times c$. We put it on an inclined plane of an angle $\alpha$ so that only one edge of length $c$ touches the plane. In time $t = 0$ the cuboid doesn't rotate. Let the line containing the edge be $k$. Let the edges of length $a$ be vertical and the ones of length $b$ - vertical.
A link to the picture is in post #1. (for the time being awaiting for mod approval)

Let the cuboid rotate around the line $k$ with angular acceleration $\varepsilon$ without sliding.

(Corrected the latex stuff)
(Based on a problem from the 58th Polish Olympiad in Physics.)

2. Relevant equations
Now it is suggested that if $a_x$ is the acceleration of the mass center parallel to the inclined plane and $a_y$ perpendicular to the plane, then
$$a_x = \varepsilon\left(\frac b2 \sin \alpha+ \frac a2 \cos \alpha \right)$$
$$a_y = \varepsilon\left(\frac b2 \cos \alpha+ \frac a2 \sin \alpha \right)$$

How can we derive it?

3. The attempt at a solution

I tried it a long time ago, using the property $a = \varepsilon r$
$$a_x = a \cos\alpha = \varepsilon r \cos\alpha$$
but from here not much success.

Last edited: Mar 20, 2015
2. Mar 20, 2015

### marcnn

3. Mar 20, 2015

### Staff: Mentor

To all: 58th Olympiad was in the 2008, so this discussion is OK.

4. Mar 22, 2015

### vela

Staff Emeritus
You seem to have a sign error somewhere. If you calculate $a_x^2+a_y^2$ from your expressions, the result depends on $\alpha$, but the magnitude of the acceleration should only depend on the shape of the block.

Try using $\vec{a} = \vec{\varepsilon} \times \vec{r}$ to calculate the acceleration of the center of mass. I wouldn't use the rotated axes to do this calculation. Then to get the component parallel and perpendicular to the incline, calculate the dot product of $\vec{a}$ with the appropriate unit vectors.

5. Mar 23, 2015

### marcnn

Of course, I made a sign mistake,

it should be
$$a_y = \varepsilon\left(\frac b2 \cos \alpha- \frac a2 \sin \alpha \right)$$

Last edited: Mar 23, 2015
6. Mar 23, 2015

### marcnn

Well, I've finally got it. It's so easy that I don't know how I could've missed it :P
It's correct, isn't it?

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