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1. What part of the region does the "plane face" refer to?

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2. (Original post by GPODT)

What part of the region does the "plane face" refer to?

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The circular face formed by rotating the line segment 360° about the x-axis.
3. (Original post by Farhan.Hanif93)
The circular face formed by rotating the line segment 360° about the x-axis.
After finding the centre of mass, they just subtracted 2 from it. So the ''plane'' face referred to was the line segment x = 2?

Also, how would one do part b? I don't usually have problems with calculating these angles but in this case I have problems visualising the shape of the circular face formed and hence I can't sketch it..

Thanks
4. (Original post by GPODT)
After finding the centre of mass, they just subtracted 2 from it. So the ''plane'' face referred to was the line segment x = 2?
The line segment x=2 is neither a plane or a face; so technically it cannot be the plane face (and it isn't) - However, when you rotate that line segment about the x-axis, you generate the plane face (rotating this entire shape all the way around the x-axis gives you a pebble-like object bounded between one flat, circular face [corresponding to rotating x=2] and a parabolical curved surface [corresponding to rotating the curved edge of the lamina]).

You subtract 2 from it still because this circular face is parallel to the x=0 plane in 3d (given that the line segment was parallel to the y-axis in 2d).

Also, how would one do part b? I don't usually have problems with calculating these angles but in this case I have problems visualising the shape of the circular face formed and hence I can't sketch it..

Thanks
Notice that this shape is rotationally symmetric about the x-axis (and is uniform) so the CoM lies on the x-axis and you can consider the toppling properties of the object by simply considering a 2d cross section (lamina) on an inclined plane instead i.e. you don't even need to visualise it in 3d beyond the rotational symmetry. Then it's a case of finding the angle at which the weight of P acts through the lower vertex.

Drawing a picture obviously helps a lot; feel free to post up a diagram if you still aren't sure about your visualisation.
5. (Original post by Farhan.Hanif93)
The line segment x=2 is neither a plane or a face; so technically it cannot be the plane face (and it isn't) - However, when you rotate that line segment about the x-axis, you generate the plane face (rotating this entire shape all the way around the x-axis gives you a pebble-like object bounded between one flat, circular face [corresponding to rotating x=2] and a parabolical curved surface [corresponding to rotating the curved edge of the lamina]).

You subtract 2 from it still because this circular face is parallel to the x=0 plane in 3d (given that the line segment was parallel to the y-axis in 2d).

Notice that this shape is rotationally symmetric about the x-axis (and is uniform) so the CoM lies on the x-axis and you can consider the toppling properties of the object by simply considering a 2d cross section (lamina) on an inclined plane instead i.e. you don't even need to visualise it in 3d beyond the rotational symmetry. Then it's a case of finding the angle at which the weight of P acts through the lower vertex.

Drawing a picture obviously helps a lot; feel free to post up a diagram if you still aren't sure about your visualisation.
Am I correct in saying that if you rotate the line segment 360 degrees about the x-axis you get a hemi-sphere like shape?

And is this a correct 2D representation? If yes, where exactly is the COM situated?

Thanks
6. (Original post by GPODT)
Am I correct in saying that if you rotate the line segment 360 degrees about the x-axis you get a hemi-sphere like shape?

And is this a correct 2D representation? If yes, where exactly is the COM situated?

Thanks
Yeah, that's close enough.

The CoM lies a distance 1.42cm from the plane face along the perpendicular that bisects the shape (which happens to be normal to the plane face).
7. (Original post by Farhan.Hanif93)
Yeah, that's close enough.

The CoM lies a distance 1.42cm from the plane face along the perpendicular that bisects the shape (which happens to be normal to the plane face).
Thanks a lot! This was very confusing but I understand now.

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