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A wheel of radius 0.4 m can rotated freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad-s^-2 is produced in it due to the torque. Then moment inertia of the wheel is (g = 10ms^-2)

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1 kg - m²

1.36 kg - m²

2 kg - m²

8 kg - m²

answer is A.

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Detailed Solution

Given,

$R = 0.4 m, α = 8 rad / s^{2}, M = 4 kg$

We know that ,

$Torque = r \times f$

$\Rightarrow I \times α = R \times F$

$I \times 8 = 0.4 \times mg$

$I \times 8 = 0.4 \times 10 \times 4$

$I = \frac{4}{10} \times 10 \times 8 \times 4 = 2 kg / m^{2}$

Hence the correct answer is $2 kg - m^{2} .$

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