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Q.

The angular velocity of a rotating disc decreases linearly with angular displacement from 60 revolutions per minute to zero during 10 revolutions as shown. The angular velocity of the disc 3 seconds after it begins to slow down is

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a

$\frac{20 π}{10}$

b

$\frac{17 π}{10}$

c

$\frac{10 π}{17}$

d

$\frac{10 π}{13}$

answer is B.

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Detailed Solution

$\begin{array}{l} As ω^{2} - ω_{0}^{2} = 2 α θ \\ \Rightarrow 0^{2} - {(2 π \times \frac{60}{60})}^{2} = 2 \times α \times (2 π \times 10) \\ \Rightarrow α = π / 10 radian / \sec^{2} \\ N o w, ω = ω_{0} + α t = 2 π \times \frac{60}{60} - \frac{π}{10} \times 3 = \frac{17 π}{10} r a d / \sec^{2} \end{array}$

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