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A homogeneous disc with a radius 0.2m and mass 5kg rotates around an axis passing through its centre. The angular velocity of rotation of the disc as a function of times is given by the formula $ω = 2 + 6 t .$ the tangential force applied to the rim of the disc is

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answer is C.

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Detailed Solution

Here, $r = 0.2 m, m = 5 kg$ As

$ω = 2 + 6 t$

$∴ α = \frac{dω}{dt} = 6 rad / s^{2}$

$τ = Iα \Rightarrow F \times r = (\frac{1}{2} {mr}^{2}) α \Rightarrow F = \frac{mrα}{2} = \frac{5 \times 0.2 \times 6}{2} = 3 N .$

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