Equilibrium of a compound object

[nokia8650]

A closed container C consists of a thin uniform hollow hemispherical bowl of radius a, together with a lid. The lid is a thin uniform circular disc, also of radius a. The centre O of the disc coincides with the centre of the hemispherical bowl. The bowl and its lid are made of the same material.

(a) Show that the centre of mass of C is at a distance from O.

The container C has mass M. A particle of mass M is attached to the container at a point P on the circumference of the lid. The container is then placed with a point of its curved surface in contact with a horizontal plane. The container rests in equilibrium with P, O and the point of contact in the same vertical plane.

(b) Find, to the nearest degree, the angle made by the line PO with the horizontal.

The solution to part b is shown below:

http://img196.imageshack.us/img196/3603/39197560.th.jpg

My question is, the solution has the line of action of the weight passing through O - why does this have to be the case; surely this assumes that O to the point of contact with the surface below is a radius on which the combined centre of mass lies - I don't see why this is essential. Thanks.

Thanks

 

[tiny-tim]

Hi nokia8650!

A closed container C consists of a thin uniform hollow hemispherical bowl
…
My question is, the solution has the line of action of the weight passing through O - why does this have to be the case

Because that's what balancing is …

the weight must go through both the c.o.m. and the point of contact.

But the c.o.m lies on a radius, and that radius must go through the point of contact, because of the geometry.

 

[nokia8650]

Thanks for the reply. Surely O - is not the point of contact - O is the centre of the upper disc.

Thanks

 

[tiny-tim]

Thanks for the reply. Surely O - is not the point of contact - O is the centre of the upper disc.

Yes, O lies on every radius …

if C is the c.o.m and P is the point of contact, then the weight must go along the line CP,

and CP must be a radius because of the geometry,

and so O also lies on CP, and the weight goes through O.