Question

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A disc of radius 0.1m rolls without sliding on a horizontal surface with a velocity of 6m/s. It then ascends a smooth continuous track. The height upto which it will ascends is​


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Answer 1

ANSWER

When the disc rolls without sliding, then the kinetic energy of the disc is,

$k = \frac{1}{2} m {v}^{2} + \frac{1}{2} i {w}^{2} = \frac{1}{2} m {v}^{2} + \frac{1}{2} \times ( \frac{1}{2} m {r}^{2} ) \frac{ {v}^{2} }{ {r}^{2} }$

K =

$\frac{1}{2} m {v}^{2} + \frac{1}{4} m {v}^{2}$

$= > \frac{3}{4} m {v}^{2}$

So, the height reached by the disc is,

So, the height reached by the disc is,mgh=34mv2

⇒h=3v24g

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