# Wobbler

A cone is glued to a hemisphere. When you place it on a table in
what position does it come to rest?

Image

A solid hemisphere of radius $a$ and a solid right circular cone of height $h$ and base radius $a$ are made from the same uniform material and joined together with their circular faces in contact. Find the position of the centre of gravity of the whole body.

If the body is placed on a horizontal table with its hemispherical surface in contact with the table in what position will it come to rest if (i) $h=a$ and (ii) $h=2a$?

(iii) If the body always rests in equilibrium when it is placed on a horizontal table with a point of the hemispherical surface in contact with the table find $h$ in terms of $a$ and the angle of the cone in this case.

(iv) In what position does the body rest in equilibrium for other angles of the cone (other values of $h/a$) and when will the equilibrium be stable?

This is all about taking moments. You could consider replacing the
hemisphere and the cone with two point masses at the centres of
mass of these two parts of the rigid body.

For different ranges of values of $h/a$ you have to consider all possible positions of equilibrium of the body and whether the positions are stable or unstable.

For different ranges of values of $h/a$ you have to consider all possible positions of equilibrium of the body and whether the positions are stable or unstable.

This solution was submitted by Andrei from Tudor Vianu National College, Bucharest, Romania.

To determine the position of the centre of gravity of the body, knowing the position of the centre of gravity for the individual bodies (i.e. the hemisphere and the circular cone), I have firstly to determine the masses of the bodies, assuming both have the same density, $\rho$.

The total mass is: $${2\over 3}\rho\pi a^3 + {1\over 3}\rho \pi a^2h.$$ I use the positions of their centres of gravity for the two distinct bodies from the hint. Taking moments about the centre of gravity of the hemisphere I obtain: $$(m_1+m_2)d = m_2\left({h\over 4} + {3a\over 8}\right)$$ where the distance $d$ is measured from the centre of gravity of the hemisphere. This gives $$\bar{x}= h + {3a^2-h^2\over 4(h+2a)}$$ from the vertex of the cone.

Label the vertex of the cone as $V$, the centre of this circular face (on the axis of symmetry) as $O$ and the point where the axis of symmetry cuts the curved surface of the hemisphere as $B$.

(i) If $h = a$, then the centre of gravity is situated in the interior of the hemisphere at a distance $a/6$ from the centre of the circular base. For this position of the centroid (and all points inside the hemisphere) the body rolls to rest with the lowest point of the hemisphere $B$ in contact with the table. The equilibrium corresponding to this position is stable, because if the body is displaced from this position, the torque of the gravitational force in respect to the point of contact with the horizontal plane, which is the point in respect to which the body could rotate, determines the body to come back to its initial position.

(ii) For $h=2a$ the centre of mass of the solid is at a distance $a/16$ from $O$ inside the cone. For this position of the centre of mass (and all positions inside the cone) the body will roll to a position of equilibrium with a generator of the cone in contact with the table. The only exceptions are when it is placed on $V$ or $B$ with the axis of symmetry vertical which are unstable positions of equilibrium.

(iii)If the body remains in equilibrium with any point of the hemisphere in contact with the plane then the centre of gravity must be at $O$ so that the centre of gravityis always vertically above the point of contact. This gives $h=a\sqrt 3$. In this vase the cross section of the cone by a plane of symmetry will be an equilateral triangle, so the angle of the cone will be $60^o$. The equilibrium is unstable when the point of contact is on the rim between the hemisphere and the cone, and stable for all other points of contact between the hemisphere and the table.

(iv) In summary

To determine the position of the centre of gravity of the body, knowing the position of the centre of gravity for the individual bodies (i.e. the hemisphere and the circular cone), I have firstly to determine the masses of the bodies, assuming both have the same density, $\rho$.

The total mass is: $${2\over 3}\rho\pi a^3 + {1\over 3}\rho \pi a^2h.$$ I use the positions of their centres of gravity for the two distinct bodies from the hint. Taking moments about the centre of gravity of the hemisphere I obtain: $$(m_1+m_2)d = m_2\left({h\over 4} + {3a\over 8}\right)$$ where the distance $d$ is measured from the centre of gravity of the hemisphere. This gives $$\bar{x}= h + {3a^2-h^2\over 4(h+2a)}$$ from the vertex of the cone.

Label the vertex of the cone as $V$, the centre of this circular face (on the axis of symmetry) as $O$ and the point where the axis of symmetry cuts the curved surface of the hemisphere as $B$.

(i) If $h = a$, then the centre of gravity is situated in the interior of the hemisphere at a distance $a/6$ from the centre of the circular base. For this position of the centroid (and all points inside the hemisphere) the body rolls to rest with the lowest point of the hemisphere $B$ in contact with the table. The equilibrium corresponding to this position is stable, because if the body is displaced from this position, the torque of the gravitational force in respect to the point of contact with the horizontal plane, which is the point in respect to which the body could rotate, determines the body to come back to its initial position.

(ii) For $h=2a$ the centre of mass of the solid is at a distance $a/16$ from $O$ inside the cone. For this position of the centre of mass (and all positions inside the cone) the body will roll to a position of equilibrium with a generator of the cone in contact with the table. The only exceptions are when it is placed on $V$ or $B$ with the axis of symmetry vertical which are unstable positions of equilibrium.

(iii)If the body remains in equilibrium with any point of the hemisphere in contact with the plane then the centre of gravity must be at $O$ so that the centre of gravityis always vertically above the point of contact. This gives $h=a\sqrt 3$. In this vase the cross section of the cone by a plane of symmetry will be an equilateral triangle, so the angle of the cone will be $60^o$. The equilibrium is unstable when the point of contact is on the rim between the hemisphere and the cone, and stable for all other points of contact between the hemisphere and the table.

(iv) In summary

- for $h< \sqrt 3$ the body comes to rest with $B$ in contact with the table and equilibrium is stable.
- for $h= \sqrt 3$ the body will rest in stable equilibrium if it is placed on the table either with a point of the hemisphere in contact with the table or with a generator of the cone in contact with the table. If it is placed with a point on the rim between the cone and the hemisphere in contact with the table then the equilibrium is semi-stable; if tipped towards the cone it will fall to rest on a generator, if tipped the other way it will rest wherever it is moved to with the hemisphere in contact with the table.
- for $h> \sqrt 3$ the body will roll to a stable equilibrium position with a generator in contact with the table. The exceptional cases are unstable equilibrium with $V$ or $B$ in contact with the table.

For this question you need to know the centres of mass of the hemisphere and the cone and they are calculated below (you will need to fill in some of the intermediate steps for yourself). You can use the positions of the centres of mass given here.

In order to find the centre of mass of the hemisphere consider a cylindrical element of thickness $\delta x$ at distance $x$ from the centre of the base $O$. Then the radius of this slice is $\sqrt{a^2 - x^2}$. The position of the centre of mass is found by taking moments about $O$ and summing these moments for all the 'slices' by integrating: $${2\over 3}w\pi a^3\bar{x}= \int_0^a x \times w\pi (a^2-x^2)dx$$ which gives $$\bar{x}={3a\over 8}.$$ Similarly to find the centre of mass of the uniform cone, a distance $\bar{x}$ along the axis of symmetry from the vertex $V$, consider cylindrical elements of thickness $\delta x$ and radius ${ax\over h}$ (from similar triangles). The position of the centre of mass is again found by taking moments, in this case about $V$: $${1\over 3}w\pi a^2h\bar{x} = \int_0^h x\times w\pi {a^2x^2\over h^2}dx$$ which gives $$\bar{x}= {3h\over 4}.$$