# Rigid bodies

#### garyljc

Question :
A uniform solid consists of a hemisphere of radius r and a right circular cone of base radius r and height h , fixed together so that their plane faces coincide . The solid can rest in equilibrium with any point of the curved surface of the hemisphere in contact with a horizontal plane.Find h in terms of r.

My working :
1. I find the mass of the cone and the hemisphere.
2. Centre of mass of the cone and the hemisphere
3. tabulate it and find the centre of the mass of the composite body.

Problem :
After find x bar of the composite body , i have no idea how to find h in terms of r . I do not understand what information they are giving me when they say that "The solid can rest in equilibrium with any point of the curved surface of the hemisphere in contact with a horizontal plane" .

Could anyone help me out . Thanks .

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#### Hootenanny

Staff Emeritus
Gold Member
garyljc said:
"The solid can rest in equilibrium with any point of the curved surface of the hemisphere in contact with a horizontal plane" .
Imagine placing this body on a table top. The object will not fall over so long as the curved surface of the hemisphere remains in contact with the table.

#### garyljc

I still dont get it . Do I use moments to find h in terms of r ?

I've already thought about it .... but then ... there's will be a lot of positions the body will be in equilibrium so as long as the curved surface is in contact with the plane . But what information does it tells us ? R=mg ?

In solving this question, you need to study the geometry of the set-up.

Consider, what is the angle between the table top and the normal reaction force on the composite body ?

Note that due to the symmetry present in the composite body, its centre of gravity must lie somewhere on the line of symmetry.

Also, for rotational equilibrium to occur, the lines of action of the weight of the body and the normal reaction force must coincide, or else a net moment will be generated.

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#### garyljc

but since the plane in horizontal , the weight will act downwards along the line of symmetry as well right ?

"but since the plane in horizontal, the weight will act downwards along the line of symmetry as well right?"

Yes, if the body is placed "vertically" on the table (i.e. with its line of symmetry perpendicular to the table).

However, analyzing such a situation only tells you that the position of the centre of gravity (CG) is SOMEWHERE on the line of symmetry. You need the exact position of the CG to be able to express h in terms of r.

To determine the exact position of the CG, consider this statement: "The solid can rest in equilibrium with any point of the curved surface of the hemisphere in contact with a horizontal plane (the table)".

Let's now consider the situation where the composite body is slanted (i.e. the line of symmetry of the body makes an angle with the table). Even in such a situation, the body can remain in equilibrium. Sketch the forces acting on the body, and remember that "for rotational equilibrium to occur, the lines of action of the weight and the normal reaction force must coincide". What can you conclude about the position of the CG?

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