What is the Height of the Cone in Terms of r?

sr-candy
Messages
15
Reaction score
0

Homework Statement


A uniform solid consist of a hemisphere of radius r and a right circular cone of base radius r fixed together so that their planes are coincident. If the solid can rest in equilibrium with any point of the curved surface of the hemisphere in contact with a horizontal plane, find the height of the cone in terms of r. [Hint: The center of gravity of the cone is 1/4 its height from the plane face, and the center of gravity of the hemisphere is 3/8 its radius from the plane face.]

Homework Equations


Xcm = (m1x1 +m2x2+...)/summation of m


The Attempt at a Solution


i tried to prove that at any point of the curved surface contacted with the ground, the volume of the left is equal to the right, but then i am having trouble in finding the volume of the varying parts.

another way that i am thinking is that, consider the CG of the cone and hemisphere separately, but again, i can't find the distance of the CGs relative to my reference point

in fact, i am not quite sure about the condition for this system to be at equilibrium

P.S. sry for my poor english as it is not my first language
 

Attachments

Physics news on Phys.org
You need to locate the center of mass, intuitively so that the object is stable on flat surfaces. And then find the max height of the cone so that it is still stable. ( Hint a hemisphere is stable because it's center of mass is lower than it's radius :P)
 
thx for replying :smile:

let the height of the cone be h and the ground be my reference point

Center of mass(CM) of the solid

CM of the hermisphere * Volume of hemishphere + CM of the cone * Volume of the cone
over the total Volume of the solid= (4/3)(r^3)(5/8r)+(r^2)(h)(h/4+r)/(4r^3+r^2h)------i canceled out "pi"

= [(5r^2)/6]+[(h^2)/4+rh]/(4/3r+h)

Am i close to the answer?

and i got three questions

1.would u mind explain more on what mean the "center of mass is lower than the radius?"

2.how can i relate the height of the cone with the radius

3.how can i show that i have consider all the cases (ie. the solid is tilted at any angle )
 
Last edited:
sr-candy said:
CM of the hermisphere * Volume of hemishphere + CM of the cone * Volume of the cone
over the total Volume of the solid

Don't quite understand that...

The center of mass multiplied to the volume?

But you need to find the height of the cone so that the center of mass of the two objects gets right between the two.That is the stability constrain.
 
nvm the problem is solved,
what i don't know before is that
if the solid can rest on any curve surface, the center of mass must lies on the contact surface of the cone and hermishpere

thx for helping anyway
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top