NEET Cut off 2024 For SC, OBC, ST & General (Reduced), Qualifying Cutoff Score For MBBS/BDS

Centre Of Mass Of Hollow Cone MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • 4 Questions around this concept.

Solve by difficulty

A hollow cone and a hollow semicircular shell are placed as shown in the diagram. Each has mass M . What is the y-coordinate of COM of system 

Concepts Covered - 1

Position of centre of mass for Hollow Cone

Have a look at the figure of Hollow Cone

  

Since it is symmetrical about y-axis  

So we can say that  its x_{cm} = 0 and z_{cm} = 0

 Now we will calculate its {y_{cm} which is given by

y_{cm} = \frac{\int y.dm}{\int dm}

So Take a small elemental ring of mass dm of radius r  at a vertical distance y from O as shown in figure.

  

And r=xsin\theta , \ and \ y=xcos\theta

Since our element mass is ring so its C.O.M will lie on the y-axis.

 Now  dm=\sigma dA=\sigma (2\pi xsin\theta )dx

Where     \sigma = \frac{M}{\pi R*\sqrt{R^2+H^2}} 

So       dm=\frac{2Mxdx}{R^2+H^2}

y_{cm}=\frac{1}{M} \int ydm=\frac{1}{M}\int_{0}^{\sqrt{R^2+H^2}}xcos\theta *\frac{2Mxdx}{R^2+H^2}=\frac{2H}{3}

 

So  \mathbf{y_{cm}= \frac{2H}{3}}  from O.

Study it with Videos

Position of centre of mass for Hollow Cone

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top