SRIHER MBBS Cut off 2024- Opening and Closing Rank

Centre Of Mass Of Semicircular Disc MCQ - Practice Questions with Answers

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

Quick Facts

  • 3 Questions around this concept.

Solve by difficulty

Two circular discs having radius \mathrm{R} and mass density \mathrm{\sigma \: and \, 2 \sigma} respectively are placed as shown in figure. The find out the position of COM will be -


 

Concepts Covered - 2

Position of centre of mass for a semicircular disc

Have a look at the figure of semicircular disc

Since it is symmetrical about y-axis on both sides of the origin

So, we can say that  its x_{cm} = 0

And its z_{cm} = 0 as  z-coordinate is zero for all particles of semicircular ring.

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 x on the disc.

dm=\frac{2M}{\pi R^2}\pi x(dx)  

As we know for semicircular ring    y_{cm}= \frac{2R}{\pi}

So, for elemental ring  y -coordinate is  y_{cm}= \frac{2x}{\pi}  

So,     y_{cm}=\frac{1}{M}\int_{0}^{R}(\frac{2x}{\pi }dm)

y_{cm}=\frac{1}{M}\int_{0}^{R}(\frac{4M}{\pi \ R^2 }x^2dx)

y_{cm}= \frac{3R}{4\pi }

Centre of mass of semicircular annular ring

Have a look at the figure of  semicircular annular ring

 

It has inner radius as R_1 and outer radius as R_2 and centre as O

Since it is symmetrical about y-axis on both sides of the origin

So we can say that  its x_{cm} = 0

And its z_{cm} = 0 as  z-coordinate is zero for all particles of semicircular ring.

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

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

So take an elemental ring of radius r  and it has mass dm and thickness as dr=dx

 

So y- coordinate of COM of an elemental ring is equal to  y_{cm_r}=\frac{2r}{\pi }

So,  y_{cm}= \frac {\int ydm}{\int dm}=\frac{\int \frac{2r}{\pi }*dm}{M}

As  \sigma =\frac{mass}{area}=\frac{M}{\frac{\pi }{2}*(R_2^2-R_1^2)}

So dm=\sigma dA=\frac{M}{\frac{\pi }{2}*(R_2^2-R_1^2)}*(\pi r*dr)

\\y_{cm}= \frac{\int \frac{2r}{\pi }*dm}{M}=\frac{\int\frac{2r}{\pi }\sigma*(\pi r*dr) }{M}=\frac{ \int 2r*\sigma *r*dr }{M}\\ \\ \\ y_{cm}=\frac{\frac{2M}{\frac{\pi }{2}*(R_2^2-R_1^2)}}{M} *\int_{R_1}^{R_2}r^2dr=\frac{4}{\pi(R_2^2-R_1^2) }*\frac{(R_2^3-R_1^3)}{3}\\ \\ \\ \\ y_{cm}=\frac{4}{3\pi }*\frac{(R_2^3-R_1^3)}{(R_2^2-R_1^2)}\\

Study it with Videos

Position of centre of mass for a semicircular disc
Centre of mass of semicircular annular ring

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

Get Answer to all your questions

Back to top