A body of mass \(5 \mathrm{~kg}\) acquires an acceleration of \(10 \mathrm{rads}^{-2}\) due to an applied…
A body of mass \(5 \mathrm{~kg}\) acquires an acceleration of \(10 \mathrm{rads}^{-2}\) due to an applied torque of \(2 \mathrm{Nm}\). Its radius of gyration is
\(2.5 \mathrm{~m}\)
\(\sqrt{2.5} \mathrm{~m}\)
\(\sqrt{0.2} \mathrm{~m}\)
\(0.2 \mathrm{~m}\)
Solution
Mass of body, \(m=5 \mathrm{~kg}\)
Angular acceleration, \(\alpha=10 \mathrm{rad} \mathrm{s}^{-2}\)
Torque, \(\tau=2 \mathrm{~N}-\mathrm{m}\)
We know that,
Torque \(=\) Moment of inertia \(\times\) Angular acceleration
\(\begin{aligned}
\Rightarrow & \tau & =I \alpha \\
\Rightarrow & I & =\frac{\tau}{\alpha}=\frac{2}{10} \\
& & =0.2 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}\)
If \(k\) be the radius of gyration, then
\(\begin{aligned}
& I=m k^2 \\
& \Rightarrow k=\sqrt{\frac{I}{m}}=\sqrt{\frac{0.2}{5}}=\sqrt{0.04}=0.2 \mathrm{~m}
\end{aligned}\)