A uniform circular disc of radius $50 c m$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of $2.0 {rads}^{- 2}$ . Its net acceleration in $m s^{- 2}$ at the end of $2 s$ is approximately

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answer is C.

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Detailed Solution

At the end of two seconds its angular velocity $ω = αt = 2 \times 2 = 4 rad / s^{2}$

Radial acceleration where $α =$ angular
acceleration $= ω^{2} r = 16 \times 0.5 = 8 m / s^{2}$ and tangential acceleration = $a_{T} = r α = 0.5 \times 2 = 1 m / s^{2}$

resultant acceleration = $\sqrt{8^{2} + 1^{2}} = \sqrt{65} \approx 8 m / s^{2}$

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