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M3: Equilibrium

A uniform thin hemispherical bowl rests in equilibrium with its axis of symmetry horizontal and its curved surface on a rough inclined plane. Show that the angle between the plane and the horizontal is 30^o.

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Okay so I modelled the bowl and a heimspherical shell, which makes the COM

\frac{r}{2}

from the centre, like so:

So I did my diagram, placing the centre of mass above the point of contact:

And that gives me

tan \theta = \frac{1}{2}

which is 27^o.

I've obviously made a mistake and I'd appreciate it if someone could tell me where :smile:

Thanks.

Reply 1

ghostwalker

Not got an easy drawing package, so a description will have to do.

So your centre of mass is above the point of contact, OK.

BUT, the perpendicular from the point of contact will go through the centre of the sphere produced, not the centre of mass.

Now draw your horizontal line from the centre of the sphere produced to the centre of gravity, and then the vertical from the centre of gravity down to the point of contact.

The hypoterneuse of the triangle produced will be r, and the distance centre of mass to centre of sphere produced r/2 as you have now, and you'll end up with 30 degrees as your angle.

Reply 2

rbnphlp

ghostwalker

OK.

BUT, the perpendicular from the point of cNoontact will go through the centre of the sphere produced, not the centre of mass.

.

sorry .would you mind expaining why this is the case? Although your approach gets the right answer ..I was told ( or read) that the point of contact goes through the line of action of the weight (COM)..

Reply 3

ghostwalker

rbnphlp

I was told ( or read) that the point of contact goes through the line of action of the weight (COM)..

It does here; I haven't said anything to contradict that.

The line of action through the centre of gravity IS NOT the perpendicular to the plane through the point of contact.

Reply 4

rbnphlp

ghostwalker

It does here; I haven't said anything to contradict that.

The line of action through the centre of gravity IS NOT the perpendicular to the plane through the point of contact.