The angular acceleration of a disc which starts rotating from rest is given by graph shown below. In which of the time intervals is the angular velocity of the disc positive.(at least for the part of a interval)

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$\frac{5 T}{2}$ to $3 T$

$T$ to $\frac{3 T}{2}$

$\frac{3 T}{2}$ to $\frac{5 T}{2}$

$3 T$ to $\frac{7 T}{2}$

answer is A, D.

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Detailed Solution

$ω_{T} = \frac{1}{2} \times (- 2 α) \times T = - v e ω_{3 T / 2} = \frac{1}{2} \times (- 2 α) \times T + \frac{1}{2} \times α \times \frac{T}{2} = - \frac{3 α T}{4} = - v e ω_{5 T / 2} = \frac{1}{2} \times (- 2 α) \times T + \frac{1}{2} \times α \times \frac{T}{2} + α \times T = \frac{α T}{4} = + v e ω_{3 T} = \frac{1}{2} \times (- 2 α) \times T + \frac{1}{2} α \times \frac{T}{2} + α T + \frac{1}{2} \times α \times \frac{T}{2} = \frac{α T}{2} = + v e ω_{\frac{7 T}{2}} = \frac{α T}{2} + \frac{1}{2} (- α) \frac{T}{2} = + v e$