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

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

Syllabus

Quick Facts

4 Questions around this concept.

Solve by difficulty

MEDIUM

A square plate of edge d and a semicircular disc of radius d are arranged as shown in figure. The centre of the combination from centre of square plate is

$\frac{9 π d - 8 d}{6 (2 + π)}$

$\frac{8 π d - 5 d}{12 π}$

$\frac{6 π d - 8 d}{8 (2 + π)}$

$\frac{3 π d}{(1 + π)}$

View Solution

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 -

$\mathrm{\frac{4 R}{3}}$

$\mathrm{\frac{2 R}{3}}$

$\mathrm{\frac{4 R}{5}}$

$\mathrm{\frac{2 R}{30}}$

View Solution

Concepts Covered - 2

Position of centre of mass for a semicircular disc

Have a look at the figure of the semicircular disc

Since it is symmetrical about the $y$ -axis on both sides of the origin
So, we can say that its $x_{c m} = 0$
And its $z_{c m} = 0_{z -coordinate is zero for all particles of the semicircular ring.}$
Now we will calculate its $y_{c m}$ which is given by

$y_{c m} = \frac{\int y \cdot d m}{\int d m}$

So, Take a small elemental ring of mass dm of radius x on the disc.

$d m = \frac{2 M}{π R^{2}} π x (d x)$

As we know semicircular ring $y_{c m} = \frac{2 R}{π}$
So, for the elemental ring $y$ -coordinate is $y_{c m} = \frac{2 x}{π}$

So,

$\begin{aligned} y_{c m} = \frac{1}{M} \int_{0}^{R} (\frac{2 x}{π} d m) \\ y_{c m} = \frac{1}{M} \int_{0}^{R} (\frac{4 M}{π R^{2}} x^{2} d x) \\ y_{c m} = \frac{3 R}{4 π} \end{aligned}$

Centre of mass of semicircular annular ring

Have a look at the figure of semicircular annular ring

It has an inner radius of $R_{1}$ an outer radius of $R_{2}$ and centre as 0
Since it is symmetrical about $y$ -axis on both sides of the origin
So we can say that its $x_{c m} = 0$
And it's $z_{c m} = 0$ as the z-coordinate is zero for all particles of the semicircular ring.
Now we will calculate its $y_{c m}$ which is given by

$y_{c m} = \frac{\int y \cdot d m}{\int d m}$

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 $y_{c m_{r}} = \frac{2 r}{π}$
${So}_{,} y_{c m} = \frac{\int y d m}{\int d m} = \frac{\int \frac{2 r}{π} * d m}{M}$
As $σ = \frac{mass}{area} = \frac{M}{\frac{π}{2} * (R_{2}^{2} - R_{1}^{2})}$
So $d m = σ d A = \frac{M}{\frac{π}{2} * (R_{2}^{2} - R_{1}^{2})} * (π r * d r)$
$y_{c m} = \frac{\int \frac{2 r}{π} * d m}{M} = \frac{\int \frac{2 r}{π} σ * (π r * d r)}{M} = \frac{\int 2 r * σ * r * d r}{M}$
$y_{c m} = \frac{\frac{2 M}{\frac{π}{2} * (R_{2}^{2} - R_{1}^{2})}}{M} * \int_{R_{1}}^{R_{2}} r^{2} d r = \frac{4}{π (R_{2}^{2} - R_{1}^{2})} * \frac{(R_{2}^{3} - R_{1}^{3})}{3}$