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M3 centre of mass watch

1. Hi guys can anyone explain why the angle marked alpha in the right hand diagram of the solution banks working is equal to the angle marked alpha on the left hand diagram.

I don’t know if there is a simple piece of geometry that explains it that i am missing. Any input would be amazing.

2. it is correct.... both of the alphas are in similar triangles... 90 degrees, alpha & ( 90 - alpha )
3. Another way of looking at it is that the left hand alpha represents the angular displacement of the plane face of the half-cylinder from the horizontal while the right hand alpha represents the angular displacement of OG (which is perpendicular to the plane face) from the vertical. Both represent the angular displacement of the object as a whole, so they must be equal.
4. (Original post by old_engineer)
Another way of looking at it is that the left hand alpha represents the angular displacement of the plane face of the half-cylinder from the horizontal while the right hand alpha represents the angular displacement of OG (which is perpendicular to the plane face) from the vertical. Both represent the angular displacement of the object as a whole, so they must be equal.
So the angle of the axis of the cylinder to the vertical is the same as the angle of the plane face to the horizontal, because the axis of the cylinder is perpendicular to the plane, and the horizontal and vertical are perpendicular so they kinda cancel out.

This makes sense.

(Original post by the bear)
it is correct.... both of the alphas are in similar triangles... 90 degrees, alpha & ( 90 - alpha )
Dont suppose u could draw out the triangles cus ive been staring at the diagram and i cant visualise the similar triangles.

5. (Original post by Shaanv)
Hi guys can anyone explain why the angle marked alpha in the right hand diagram of the solution banks working is equal to the angle marked alpha on the left hand diagram.

I don’t know if there is a simple piece of geometry that explains it that i am missing. Any input would be amazing.

There is a much easier explanation as to why the angle is alpha... have a look below. If you look at the quadrilateral OTSG you know the angle TSG and TOG must be 90 degrees as GS is vertical and TS is horizontal ( center of mass must lie directly above the contact point S so there is no resultant moment and hence prism is in equilibrium) and OG is perpendicular to the rectangular plane. That means the remaining two angles must add to make 180 degrees (sum of angles in quadrilateral is 360 degrees). You can therefore deduce that the mystery angle is also alpha and proceed with the question
6. (Original post by Anonymouspsych)
There is a much easier explanation as to why the angle is alpha... have a look below. If you look at the quadrilateral OTSG you know the angle TSG and TOG must be 90 degrees as GS is vertical and TS is horizontal ( center of mass must lie directly above the contact point S so there is no resultant moment and hence prism is in equilibrium) and OG is perpendicular to the rectangular plane. That means the remaining two angles must add to make 180 degrees (sum of angles in quadrilateral is 360 degrees). You can therefore deduce that the mystery angle is also alpha and proceed with the question
You're assuming that a horizontal line drawn rightwards from T passes through the point of contact S. This will not generally be the case.
7. (Original post by old_engineer)
You're assuming that a horizontal line drawn rightwards from T passes through the point of contact S. This will not generally be the case.
It doesn't have to pass through the contact point though surely- no matter how the cylinder is positioned to achieve equilibrium if you think about it the rightward horizontal line from one of the vertex will always perpendicularly cut the line of action of the centre of weight and hence you will always be able to draw a similar quadrilateral ?? As long as the horizontal line doesn't cross above the centre of mass? Correct me if I'm wrong

Edit: Ahhhh nvm I see your point
8. (Original post by Shaanv)
So the angle of the axis of the cylinder to the vertical is the same as the angle of the plane face to the horizontal, because the axis of the cylinder is perpendicular to the plane, and the horizontal and vertical are perpendicular so they kinda cancel out.

This makes sense.

Dont suppose u could draw out the triangles cus ive been staring at the diagram and i cant visualise the similar triangles.

Hey, please ignore my previous solution. Thanks to old_engineer I realised that I have made a massive assumption- below is the proper explanation. It shows why the angle is alpha even when the horizontal line doesn't pass through the contact point S. Pay attention to the triangle I have marked as red you can see there is a smaller triangle within the bigger red triangle and they are clearly similar triangles and this leads to the mystery angle being alpha . Hope this helps
9. (Original post by Anonymouspsych)
Hey, please ignore my previous solution. Thanks to old_engineer I realised that I have made a massive assumption- below is the proper explanation. It shows why the angle is alpha even when the horizontal line doesn't pass through the contact point S. Pay attention to the triangle I have marked as red you can see there is a smaller triangle within the bigger red triangle and they are clearly similar triangles and this leads to the mystery angle being alpha . Hope this helps
Haha i knew it was a simple bit of geometry. Completely missed the similar triangles, thanks for clarifying with a diagram

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Updated: October 22, 2017
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