M3 exam Q

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


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The circular face formed by rotating the line segment $x=2, 0\leq y\leq 4$ 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

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] 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

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).