Centre of mass of a hollow cone whose mass per unit area is fixed
Answers
Seeing this question I guess you are having problem understanding basic concepts of centre of mass. Therefore, I won’t answer your question! Instead of that I will just give you the hints to do it and try to figure out on yourself. Giving the answer to you won’t help you much until you can do it on your own.Since you haven’t mentioned if it is a hollow or a solid cone I will just tell you for solid cone.
STEP 1 : Make the diagram of the cone( radius=R,height=H,volume=V and mass= M) and create an elemental disk of mass “dm”, radius “r” and width dy on it.
STEP 2 : Since it is a uniform cone (assumed), It has a uniform density.Hence the mass dm of the elemental disk will be ( dm= (M/V) dV ), here dV=(
pi∗
r
2
)dy.
pi∗r2)dy.
STEP 3 : There you can see are two triangle which are similar, therefore we can say
r/h=R/H. Using this you can put value of r in terms of R,H and y. Some terms will cancel out and you will get dm.
Step 4: Use formula for y-coordinate ( since it will be on y-axis) , integrate [(1/M) dm.y]. SImplify and you’ll get the answer. And remember use the limits from 0 to H since the disk will go all the way from top to bottom in order to integrate.
Try for yourself and if you have problems anywhere ask me in the comments I will be happy to help you. And match the answer ( It’s 3H/4). If you want to know for hollow cone keep the steps same and just find dm= (M/V)dA ( it has area only) and dA=(
2∗pi∗r)dy
2∗pi∗r)dy
. Do the same things after that, integration and all stuff.
PS: I am not good at explaining things so, sorry if I couldn’t explain it good.
the image is not getting uploaded. It might have helped. I’ll try it later.