TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

The situation given can be shown as. Here, R be the radius of circular disc =2cm
The radius of removed portion, r=1cm
Area of whole disc, A1=πR2=4πcm2
Area of removed portion, A2=πr2=πcm2
As, ‌

1

4

th area of the disc is removed, so remaining
area of the disc is ‌

3

4

th initial area.
Similarly, remaining mass, m1=‌

3

4

m
Mass of removed portion, m2=‌

1

4

m
Let x be the maximum shift in centre of mass. So, initially centre of mass along X -axis of complete disc,
xCM=‌

m1x+m2r

m1+m2

0=‌

‌ 3 4 mx+ 1 4 mr

3 4 m+‌ 1 4 m

⇒‌‌‌

3

4

mx=−‌

1

4

mr or x=−‌

1

3

r=−‌

1

3

cm
i.e., the centre of mass of remaining portion will shift to the left of origin at ‌

1

3

cm.
Now, the disc is rotated by angle θ, so the centre of mass will also shift by angle θ as shown
∆OPQ is isoscales. So, a perpendicular drawn from 0, on PQ divide the angle θ and length PQ in equal parts i.e.,
RQ=‌

1

2

PQ=‌

1

2

(‌

1

√3

)cm‌‌( given, PQ=‌

1

√3

)
From ∆ORQ.
sin‌

θ

2

=‌

RQ

OQ

=‌

‌ 1 2 × 1 √3

1 3

⇒‌‌sin‌

θ

2

=‌

√3

2

=sin‌60∘
⇒‌‌θ=120∘