TG EAMCET 3 May 2025 Shift 2 Paper

Moment of inertia of ring about a tangential axis in its plane
I=ICM+MR2=‌

MR2

2

+MR2
[The moment of inertia of a ring about its diameter ICM=‌

MR2

2

]
I=‌

3

2

MR2
If KR be the radius of gyration, then
MKR2=‌

3

2

MR2⇒KR=√‌

3

2

R
Moment of inertia of circular disc about tangential axis in its plane
I‌=ICM+MR2
‌=‌

MR2

4

+MR2=‌

5

4

MR2
[The moment of inertia of a disc about its diameter is ICM=‌

1

4

MR2 ]
If KC be the radius of gyration, then
MKC2‌=‌

5

4

MR2
KC‌=√‌

5

4

R
‌ Since, ‌‌

KR

KC

‌=‌

√12

√K

‌

√ 3 2 R

√‌ 5 4 R

‌=‌

√12

√K

⇒‌

3

2

×‌

4

5

=‌

12

K

⇒‌‌K‌=10