The moment of inertia of a wheel of radius 20 cm is 40kg m2 . If a tangential force of 80 N applied on the wheel. Its rotational K.E. after 4s is

τ=rF=0.2×80=16N−mα=τI=1640=25rad/s2KE gained in 4s is E=12Iω2=12I(αt)2=12(40)25×42=51.2J

The moment of inertia of a wheel of radius 20 cm is 40kg m2 . If a tangential force of 80 N applied on the wheel. Its rotational K.E. after 4s is

τ=rF=0.2×80=16N−mα=τI=1640=25rad/s2KE gained in 4s is E=12Iω2=12I(αt)2=12(40)25×42=51.2J

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The moment of inertia of a wheel of radius 20 cm is 40kg m² . If a tangential force of 80 N applied on the wheel. Its rotational K.E. after 4s is

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16.2 J

51.2J

25.6J

24.8J

answer is B.

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Detailed Solution

$\begin{array}{l} τ = rF = 0 .2 \times 80 = 16 N - m \\ α = \frac{τ}{I} = \frac{16}{40} = \frac{2}{5} rad / s^{2} \end{array}$
KE gained in 4s is $E = \frac{1}{2} I ω^{2} = \frac{1}{2} I (αt)^{2} = \frac{1}{2} (40) {(\frac{2}{5} \times 4)}^{2} = 51 .2 J$