Prove that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of height h is given by:$\text{v}^2=\frac{2\text{gh}}{\big(1+\frac{\text{K}^2}{\text{R}^2}\big)}$
Note K is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
Note K is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
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$\text{Ma}=\text{Mg}\sin\theta-\text{f}\dots(1)$$\tau=\text{f}\times\text{R}=\text{ I }\alpha=\text{MK}^2\Big(\frac{\text{a}}{\text{R}}\Big)$
$\Rightarrow\text{f}=\frac{\text{MK}^2}{\text{R}^2}\text{a}$
$\Rightarrow\text{Ma}=\text{Mg}\sin \theta-\frac{\text{MK}^2}{\text{R}^2}\text{a}$
[Substituting the value of f in (i)]
$\Rightarrow\text{a}\Big(1+\frac{\text{K}^2}{\text{R}^2}\Big)=\text{g}\sin\theta$
$\Rightarrow\text{a}=\frac{\text{g}\sin\theta}{\Big(1+\frac{\text{K}^2}{\text{R}^2}\Big)}$
$\text{v}^2-\text{u}^2=2\text{as}[\text{u}=0,\frac{\text{h}}{\text{s}}=\sin\theta]$
$\text{v}=\sqrt{\frac{2\text{ah}}{\sin\theta}}$
$\Rightarrow\text{v}=\sqrt{\frac{2\times\text{g}\sin\theta}{\sin\theta\Big(1+\frac{\text{K}^2}{\text{R}^2}\Big)}}$
$\Rightarrow\text{v}=\sqrt{\frac{2\text{gh}}{\Big(1+\frac{\text{K}^2}{\text{R}^2}\Big)}}$
Where, K = radius of gyration of the body about its symmetry axis.
R = Radius of the body.
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