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The moment of inertia of a solid flywheel about its axis is $0.1\,kg-m^2$. A tangential force of $2\,kg\,wt$. is applied round the circumference of the flyweel with the help of a string and mass arrangement as shown in the figure. If the radius of the wheel is $0.1\,m,$ find the angular acceleration of the solid fly wheel (in $rad/sec^2$)
$163.3$
$16.3$
$81.66$
$8.16$
Solution
Suppose a be the linear acceleration of the mass and $T$ the tension in the string.
Hence, $\quad \mathrm{Mg}-\mathrm{T}=\mathrm{Ma}$ $…(1)$
Let $\alpha$ be the angular acceleration of the flywheel. The couple applied to the flywheel is
$\mathrm{I} \alpha=\mathrm{TR} \quad$ or $\quad \mathrm{T}=\frac{\mathrm{I} \alpha}{\mathrm{R}}$ $…(2)$
Now we know that $a=R \alpha$ $…(3)$
Putting $( 2)$ and $( 3)$ in eq. $(1),$ we get
$\mathrm{Mg}-\frac{\mathrm{I} \alpha}{\mathrm{R}}=\mathrm{MR} \alpha$
$\therefore \quad \alpha=\frac{\mathrm{MgR}}{\mathrm{I}+\mathrm{MR}^{2}}$ $…(4)$
$=\frac{2 \times 9.8 \times 0.1}{0.1+2 \times(0.1)^{2}}=16.33 \mathrm{rad} / \mathrm{sec}^{2}$
