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

The disk of radius 5 cm is cut from the uniform disc of radius 10 cm in such a way that the edge of the hole touches the edge of the disc. What is the distance of the center of mass of the remaining portion w.r.t the point O ?


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

The correct option is C 1.66 cm

Let ρ be the mass density of the disk.
Let m1 be the mass of circular disk of radius 10 cm and m2 be the mass of circular disk of radius 5 cm.
Take O to be the origin for reference.
So, x1=0 (COM of circular disk of radius 10 cm)
x2=5 (COM of removed disk).

Now,
XCOM=m1x1m2x2m1m2
(we treat the removed portion of disc as having negative mass)
where
m1=ρ(π(10)2)=100πρ
m2=ρ(π(5)2)=25πρ
XCOM=100πρ×(0)25πρ×(5)100πρ25πρ=25×510025
=1.66
Due to symmetry, YCOM will be zero.

Therefore, (XCOM,YCOM)=(1.66,0).
So, COM of remaining portion will be shift towards left side on the line joining the points 'O' and 'C'.

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon