TS EAMCET 04 May 2019 Shift 2 Solved Paper

It is given that
angular acceleration ∝√‌ angualar speed ‌

The above expression can be rewritten as,
‌

−dω

dt

∝√ω
Add a constant kto remove proportionality sign. ‌

−dω

dt

=k√ω
Integrate the above equation.

−

∫

ω0

ω‌

dω

√ω

=k‌

t

∫

0

dt
−[2√ω]ω0ω=k[t]0t
√ω−√ω0=−‌

kt

2

Substitute 0 for ω in the above equation.

t=‌

2√ω0

k

Therefore, mean angular speed is given by,
⟨ω⟩‌‌=‌

∫ 0 2 √ω0 k (ω0+ k2t2 4 −kt√ω0)dt

2 √ω0 k

‌‌=‌

[(ω0t+ k2t3 12 − k 2 √ω0t2)]02 √ω0 k

2 √ω0 k

‌‌=‌

ω0

3