wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

From a uniform circular disc of radius 20 cm, two small circular discs of radius 10 cm are removed in such a way that both have common tangent as shown in the figure given below. The distance of centre of mass of the remaining part from the centre of original disc is:


A
10 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0 cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
5 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
20 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 0 cm
Given radius of original disc, R=20 cm
Let, mass of original disc be M
Let the mass density of the disc be σ
σ×(πR2)=M
σ=MπR2

Mass of removed disc is σ×πR24=M4
Both removed discs have mass M4 each.
Let COM of the original disc be at the origin.


x-Coordinate of COM of the remaining part is given by :-
xCOM=M(0)M4(R2)M4(R2)MM4M4
=0MR8+MR8MM2=0
xCOM=0
Similarly,
y-coordinate of COM of remaining portion will be the same that of original disc.
yCOM=0

i.e center of mass of the remaining part lies at the center of original disc.
So, the distance of the centre of mass of the remaining part from the center of mass of the original disc is 0.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon