A wheel of radius \(8 \mathrm{~cm}\) is attached to a support so as to rotate about a horizontal axis…
A wheel of radius \(8 \mathrm{~cm}\) is attached to a support so as to rotate about a horizontal axis through its centre. A string of negligible mass wrapped around its circumference carries a mass of \(0.4 \mathrm{~kg}\) attached to its free end. When the mass is released, it descends through \(1 \mathrm{~m}\) in 10 seconds, then its moment of inertia is (Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\))
\(1.277 \mathrm{~kg} \mathrm{~m}^2\)
\(2.177 \mathrm{~kg} \mathrm{~m}^2\)
\(21.77 \mathrm{~kg} \mathrm{~m}^2\)
\(12.77 \mathrm{~kg} \mathrm{~m}^2\)
Solution
Given, radius of wheel, \(R=8 \times 10^{-2} \mathrm{~m}\), mass of weight, \(m=0.4 \mathrm{~kg}\), descending length, \(L=1 \mathrm{~m}\) and time, \(t=10 \mathrm{~s}\)
As, torque \(\tau=F . R=m g R\) putting the given values, we get
\(=0.4 \times 10 \times 8 \times 10^{-2}\)
So, \(\quad \tau=0.32 \mathrm{Nm}\)
Let the wheel is rotated by an angle \(\theta\),
\(\theta=\frac{\text { arc }}{\text { radius }}=\frac{1}{0.08}=12.5 \mathrm{rad}\)
So, from the equation of rotational motion,
\(\begin{aligned}
\quad \theta & =\omega_0 t+\frac{\alpha t^2}{2} \\
\Rightarrow \quad \alpha & =\frac{2 \theta}{t^2}=\frac{2 \times 12.5}{10 \times 10}=0.25 \mathrm{rad} \mathrm{s}^{-2}
\end{aligned}\)
\(\left[\because \omega_0=0\right]\)
Now, as
Torque, \(\quad \tau=I \alpha\)
\(\therefore \quad\) Moment of inertia of a wheel,
\(I=\frac{\tau}{\alpha}=\frac{0.32}{0.25}=1.28 \mathrm{~kg}-\mathrm{m}^2\)
Hence, the correct option is (a).
