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A wheel of radius 0.4 m can rotate 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 $8 \mathrm{~rad/s}^{-2}$ is produced in it due to the torque. Then, moment of inertia of the wheel is $\left(g=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$2 \mathrm{~kg}-\mathrm{m}^2$
Given, $\quad r=0.4 \mathrm{~m}$
$\alpha=8 \mathrm{~rad}/ \mathrm{s}^{-1}$
$m=4 \mathrm{~kg}, \quad I=$ ?
Torque, $\quad \tau=I \alpha$
$m g r=I \cdot \alpha$
or $\quad 4 \times 10 \times 0.4=I \times 8$
$\Rightarrow \quad I=\frac{16}{8}=2 \mathrm{~kg}-\mathrm{m}^2$
or $I=2 \mathrm{~kg}-\mathrm{m}^2$
$\alpha=8 \mathrm{~rad}/ \mathrm{s}^{-1}$
$m=4 \mathrm{~kg}, \quad I=$ ?
Torque, $\quad \tau=I \alpha$
$m g r=I \cdot \alpha$
or $\quad 4 \times 10 \times 0.4=I \times 8$
$\Rightarrow \quad I=\frac{16}{8}=2 \mathrm{~kg}-\mathrm{m}^2$
or $I=2 \mathrm{~kg}-\mathrm{m}^2$
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