wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A circular plate of uniform thickness has a diameter of 28 cm. A circular portion of diameter 21 cm is removed from the plate as shown. Take the origin at the center of mass of the complete plate. The position of the center of mass of the remaining portion will shift towards left from the origin by


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

The correct option is C 4.5 cm

Let x1 be the position of the COM of the removed portion and x2 be the position of the center of mass of the remaining portion. (taking origin at point O)
Due to symmetry, YCOM will be zero.

Area of circular portion which is removed A1=π(10.5)2
Area of portion remaining A2=π(14210.52)

Hence, XCOM=A1x1+A2x2A1+A2
Here, x1=1410.5=3.5 cm

Since centre of mass of the complete plate is at the origin
0=π(10.5)2×3.5+π(14210.52)x2A1+A2
x2=(10.5)2×3.514210.52=4.5 cm
x2=4.5 cm {-ve sign is showing it is towards left of origin }

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integrating Solids into the Picture
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon