A fly-wheel of mass $25 \mathrm{~kg}$ has a radius of $0.2 \mathrm{~m}$. It is making $240 \mathrm{rpm}$.…

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 $20 \mathrm{~s}$ ?
  1. $2 \pi \mathrm{Nm}$
  2. $0.4 \pi \mathrm{Nm}$
  3. $\frac{2}{\pi} \mathrm{Nm}$
  4. $4 \pi \mathrm{Nm}$

Solution

$\alpha=\frac{2 \pi n}{t}=\frac{2 \pi \times \frac{240}{60}}{20}$ $=\frac{2 \pi \times 4}{20}$ $\alpha=\frac{2 \pi}{5}$ Torque, $\quad \tau=I \alpha$ $\tau=M R^2 \alpha$ $=25 \times(0.04) \times \frac{2 \pi}{5}$ $=0.4 \pi \mathrm{Nm}$

Asked in: AP EAMCET 2010

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