Prove the result 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 a height h is given by $\text{v}^2=\frac{2\text{gh}}{\Big(\frac{1+\text{k}^2}{\text{R}^2}\Big)}$ using dynamical consideration (i.e. by consideration of forces and torques).
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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A body rolling on an inclined plane of height h,is shown in the following figure:
m = Mass of the body. R = Radius of the body. K = Radius of gyration of the body. v = Translational velocity of the body. h =Height of the inclined plane. g = Acceleration due to gravity. Total energy at the top of the plane, $E_1$ = mgh Total energy at the bottom of the plane, $E_b = KE_{rot} + KE_{trans} $
$=\frac{1/2}{\omega^2}+\frac{1}{2}\text{mv}^2$ But $I = mk^2$ and $\omega=\frac{\text{v}}{\text{R}}$ $\therefore\ \text{E}_{\text{b}}=\frac{1\text{mk}^2\text{v}^2}{2\text{R}^2}+\frac{1}{2}\text{mv}^2$ From the law of conservation of energy, we have, $\text{E}_{\text{T}}=\text{E}_{\text{b}}$ $\text{mgh}=\Big(\frac{1\text{mv}^21+\text{k}^2}{2\text{R}^2}\Big)$
$\therefore\ \text{v}^2=\frac{2\text{gh}}{\Big(\frac{1+\text{k}^2}{\text{R}^2}\Big)}$ Hence, the given result is proved.
m = Mass of the body. R = Radius of the body. K = Radius of gyration of the body. v = Translational velocity of the body. h =Height of the inclined plane. g = Acceleration due to gravity. Total energy at the top of the plane, $E_1$ = mgh Total energy at the bottom of the plane, $E_b = KE_{rot} + KE_{trans} $
$=\frac{1/2}{\omega^2}+\frac{1}{2}\text{mv}^2$ But $I = mk^2$ and $\omega=\frac{\text{v}}{\text{R}}$ $\therefore\ \text{E}_{\text{b}}=\frac{1\text{mk}^2\text{v}^2}{2\text{R}^2}+\frac{1}{2}\text{mv}^2$ From the law of conservation of energy, we have, $\text{E}_{\text{T}}=\text{E}_{\text{b}}$ $\text{mgh}=\Big(\frac{1\text{mv}^21+\text{k}^2}{2\text{R}^2}\Big)$
$\therefore\ \text{v}^2=\frac{2\text{gh}}{\Big(\frac{1+\text{k}^2}{\text{R}^2}\Big)}$ Hence, the given result is proved.
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