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A fly-wheel of mass $25 \mathrm{~kg}$ has a radius of $0.2 \mathrm{~m}$. It is making $240 \mathrm{rpm}$. What is the torque necessary to bring to rest in $20 \mathrm{~s}$ ?
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Verified Answer
The correct answer is:
$0.4 \pi \mathrm{Nm}$
$\alpha=\frac{2 \pi n}{t}=\frac{2 \pi \times \frac{240}{60}}{20}$
$=\frac{2 \pi \times 4}{20}$
$\alpha=\frac{2 \pi}{5}$
Torque, $\quad \tau=I \alpha$
$\tau=M R^2 \alpha$
$=25 \times(0.04) \times \frac{2 \pi}{5}$
$=0.4 \pi \mathrm{Nm}$
$=\frac{2 \pi \times 4}{20}$
$\alpha=\frac{2 \pi}{5}$
Torque, $\quad \tau=I \alpha$
$\tau=M R^2 \alpha$
$=25 \times(0.04) \times \frac{2 \pi}{5}$
$=0.4 \pi \mathrm{Nm}$
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