TS EAMCET 04 May 2019 Shift 2 Solved Paper

Consider the figure shown below.
The linear acceleration of a solid sphere is given by,
a=‌

g‌sin‌θ

1+‌ I MR2

=‌

g‌sin‌θ

1+‌ ‌ 2 5 MR2 MR2

=‌

5

7

g‌sin‌θ
Apply Newton's second law of motion for linear motion of the centre of mass.
mg‌sin‌θ−f‌‌=ma
f‌‌=mg‌sin‌θ−ma
‌‌=mg‌sin‌θ−‌

5

7

mg‌sin‌θ
‌‌=‌

2

7

mg‌sin‌θ
The coefficient of friction is calculated as,
(µS)=‌

f

N

=‌

‌ 2 7 mg‌sin‌θ

mg‌cos‌θ

=‌

2

7

‌tan‌θ
To prevent slipping,
µs≥‌

2

7

‌tan‌θ
≥‌

2

7

‌tan‌45∘
≥‌

2

7

Hence, ‌

3

7

,‌

1

2

and ‌

5

8

are greater than ‌

2

7