The moment of inertia of a solid flywheel about its axis is $0.1kg-m^2$. A tangential force of $2kg/ wt$ is applied round the circumference of the flywheel with the help of a string and mass arrangement as shown in Fig. If the radius of the wheel is $0.1m$, find the acceleration of the mass.
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Let a be the linear acceleration of the mass and T the tension in the spring. It is clear that $\text{mg}-\text{T}=\text{ma}\dots(1)$ Let the angular acceleration of the flywheel be α. The couple applied to the flywheel is $\text{l}\alpha=\text{TR}\dots(2)$ The linear acceleration a and angular acceleration are related to each other as $\text{r}=\text{R}\ \alpha\dots(3)$ Combining Eq. (1), (2) and (3), we get $\text{mg}-\frac{\text{l}\alpha}{\text{R}}=\text{m}\ \text{R}\ \alpha$ $\alpha=\frac{\text{m}_{\text{g}}\text{R}}{(1+\text{m}\ \text{R})^2}\dots(4)$ It is given that $= 2kg, R = 0.1m$ and $I = 0.1kgm^2$.Substituting these values, We get $\alpha=\frac{2\times9.8\times0.1}{(0.1+2\times0.1^2)}\text{rad}\ \text{s}^{-2}$ $=16.7\ \text{rad}\ \text{s}^{-2}$
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