Center of Mass in a system of cylinders

AI Thread Summary
The discussion focuses on calculating the center of mass (COM) for a composite system consisting of a brass collar and an aluminum bar. The correct formula for COM is provided, but the user struggles with their calculations, particularly regarding the mass and dimensions of the components. There is confusion about subtracting the mass of a collar hole, which is deemed unnecessary if the volume was already calculated correctly. Participants emphasize the importance of using the correct reference points for the COM calculations, especially considering the symmetry of the objects involved. The goal is to arrive at a COM value of 27.6 mm based on accurate calculations.
noffya
Messages
5
Reaction score
0

Homework Statement



A collar brass 50 mm length is mounted on an aluminum bar 80 mm in length (density of brass 8470 kg / m 3 density of aluminum 2800 kg / m 3 ). Find the height to which is the center of mass of the composite body.

vssY5I9kHdQ.jpg


Homework Equations



COM= m1*x1+m2x2/m1+m2


The Attempt at a Solution


ZHMt7UdftMU.jpg




I tried to solve the problem using the coordinates x,y,z, however didn't get a right solution.
The solution must be equal 27.6 mm
Please help with the equations.
Thanks
 
Last edited:
Physics news on Phys.org
You can treat this as a 1d problem. Work out the centre of mass of the two objects. You can then work out the centre of mass of the two points by using the mass of the two objects to weight them.
 
If your picture was a little smaller, you couldn't tell what you were doing.

It looks like you calculated the volume of the collar and the bar correctly. What I don't understand is why you have subtracted a mass of 0.2831 kg from the masses of the collar and the bar.

Also, the x values in the equation for the c.o.m. are not the heights of the collar or the bar; the x-values are the x locations of the c.o.m. for each item from the reference.

Since the collar and the bar are symmetrical w.r.t. the y and z axes, your values for the c.o.m. w.r.t. these axes are incorrect. The c.o.m. for a uniformly distributed mass will lie on any axes of symmetry which the mass may have.
 
noffya said:

Homework Statement



A collar brass 50 mm length is mounted on an aluminum bar 80 mm in length (density of brass 8470 kg / m 3 density of aluminum 2800 kg / m 3 ). Find the height to which is the center of mass of the composite body.

vssY5I9kHdQ.jpg


Homework Equations



COM= m1*x1+m2x2/m1+m2


The Attempt at a Solution


ZHMt7UdftMU.jpg




I tried to solve the problem using the coordinates x,y,z, however didn't get a right solution.
The solution must be equal 27.6 mm
Please help with the equations.
Thanks
Your handwriting is unreadable. Please type out the relationships and results you got for the volumes of the two objects and the masses of the two objects.
Chet
 
SteamKing said:
If your picture was a little smaller, you couldn't tell what you were doing.

It looks like you calculated the volume of the collar and the bar correctly. What I don't understand is why you have subtracted a mass of 0.2831 kg from the masses of the collar and the bar.

Also, the x values in the equation for the c.o.m. are not the heights of the collar or the bar; the x-values are the x locations of the c.o.m. for each item from the reference.

Since the collar and the bar are symmetrical w.r.t. the y and z axes, your values for the c.o.m. w.r.t. these axes are incorrect. The c.o.m. for a uniformly distributed mass will lie on any axes of symmetry which the mass may have.


thanks a lot!
I subtracted a mass of 0,2831kg considering it a collar hole.
I will try to make calculation based on all the comments. Hopefully that will lead me to the right solution.
 
noffya said:
thanks a lot!
I subtracted a mass of 0,2831kg considering it a collar hole.
I will try to make calculation based on all the comments. Hopefully that will lead me to the right solution.

There is no need to subtract the mass of the collar hole if you calculated the volume of a cylinder with a hole removed in the first place, which it appears you did.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top