CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A circular disc of radius 2 m has a hole of radius 1 m at its centre. Then, find the radius of gyration of the disc about the axis passing through its centre and perpendicular to its plane. Given that mass per unit area of the disc varies as σ0r.


A
213 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
313 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
203 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
413 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 213 m
Let us take a small elemental ring of mass dm, radius r and width dr as shown in figure.


Moment of inertia of ring
dI=r2dm
dI=r2σ02πdr ...(i)
[Given, mass per unit area varies as σ0r,
For the elemental ring,
mass =dm, area =2πrdr
dm2πrdr=σ0r ...(ii)]
Integrating Eq.(i) with limits of radius r=1 m2 m
I0dI=2σ0π21r2dr
I=2σ0π[r33]21
I=14σ0π3
For finding radius of gyration about the axis through the centre of the disc,
I=mK2
Putting the value of I, we get
mK2=14σ0π3 ...(iii)
Since, m=m0dm and substituting the value of dm from Eq.(ii)
m0dm=r=2r=1σ02πrdrr
m=2πσ0

Putting in Eq.(iii), we get,
2πσ0K2=14σ0π3
K=73
K=213 m
The radius of gyration of the disc about its axis passing through the centre is 213 m.

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