Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

Mass per unit of a semicircular disc of total mass m radius R varies linearly with radial distances from the point 0(base centre). Find the height of the centre of mass from 0.
A2RπB3R2πC4R3πD3R8

Answer
VerifiedVerified
439k+ views
2 likes
like imagedislike image
Hint: Centre of mass of an object is the point at which an object can be balanced and its position depends on the shape of the body as it is present at a point nearer to the more mass concentrated area.

Complete step by step answer:
Now we have to find out height of centre of mass from the point 0,
seo images

                                                 Fig. a
For this, first consider a semicircular disc of radius ‘R’ with base ‘0’ as shown in the above figure, whose mass is ‘M’ distributed uniformly over the disk.
Now, to find the centre of mass of a semicircular disc, it can be divided into several rings. Now consider one such ring of thickness ‘dr’ and radius ‘r’ and mass ‘dm’ as shown in figure a.
If we consider a circular disk, then its centre of mass lies at the centre of the disk but when we consider a semicircular disk its centre of mass shifted to above in the y-axis as shown.
So by symmetry centre of mass lie on y-axis , i.e., xcm=0 so there exist only Ycm
The centre of mass formula of a continuous body is given by:
Ycm=ydmM …….. equation (1)
Where, y is the centre of mass of the ring,
y=2rπ where r is the radius of the small ring.
M is the total mass of the disk
And dm is the mass of the ring
To substitute dm (mass of the small ring) in the above formula, first we have to find it.
Let us consider ring in the form of stripe as shown below:
seo images

                                           Fig. b
now we know that, to find the mass of the ring the formula is:
dm=MAdA …….. equation (2)
where, M - is the total mass of semicircular disc
A – is the area of the total semicircular disc
dA - is the area of the considered ring

Now area of small ring = length × width
 width is ‘dr’ as shown since it is a semicircle its length ‘l’ is half of the circumference of the circle,
l=2πr2=πr
On substituting we get,
 dA=πrdr
on substituting above values in equation 2, we get,
dM=Mπr22dA
dM=2MrdrR2
Thus, we find the mass of the ring.
Now on substituting all the values including dm in equation (1), we get,
Ycm=2r2MrdrπR2M
Take constants out of the integral and cancelling M, we get,
Ycm=4πR20Rr2dr
Limits are taken from 0 to R because we measure it from centre of the semicircle to total radius R up to edge covering the entire semicircular disc
on integrating we get,
Ycm=4πR2[r33]0R

substituting the limits, we get,
Ycm=4πR2[R33]
As R is in both numerator and denominator by cancelling, we get
YCM=4R3π
Thus, the height of the centre of mass from base ‘0’ is 4R3πwhich means the correct option is C.

Note: Students make mistakes by confusing it with a ring and may choose option A, which is incorrect. They have asked for a disk where mass is distributed continuously. Always read questions properly.
To solve this problem you have to remember all the above formulas along with the centre of mass of the semicircular ring which is =2Rπ.
Latest Vedantu courses for you
Grade 10 | MAHARASHTRABOARD | SCHOOL | English
Vedantu 10 Maharashtra Pro Lite (2025-26)
calendar iconAcademic year 2025-26
language iconENGLISH
book iconUnlimited access till final school exam
tick
School Full course for MAHARASHTRABOARD students
PhysicsPhysics
BiologyBiology
ChemistryChemistry
MathsMaths
₹36,600 (9% Off)
₹33,300 per year
Select and buy