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The moment of inertia of a wheel of radius 20 cm is $40 k g m^{2}$ . If a tangential force of 80 N applied on the wheel, then its rotational K.E. after 4s is

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16.2 J

24.8 J

51.2 J

25.6 J

answer is B.

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Detailed Solution

$τ = r F = 0.2 \times 80 = 16 N - m$

$α = \frac{τ}{I} = \frac{16}{40} = \frac{2}{5} r a d / s^{2}$

KE gained in 4s is $E = \frac{1}{2} I w^{2} = \frac{1}{2} I {(α t)}^{2} = \frac{1}{2} (40) {(\frac{2}{5} \times 4)}^{2} = 51.2 J$

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