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  4. A disc of radius text0.1 m rolls without sliding on a horizontal surface with a velocity of 6 ms- 1 . It then ascends a smooth continuous track as shown in figure. The height upto which it will ascend is (g = 10 m s- 2) <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-vzdsomhhym9f.png />

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Q. A disc of radius $\text{0.1 m}$ rolls without sliding on a horizontal surface with a velocity of $6 \, ms^{- 1}$ . It then ascends a smooth continuous track as shown in figure. The height upto which it will ascend is $\left(g = 10 m s^{- 2}\right)$

Question

NTA AbhyasNTA Abhyas 2020Work, Energy and Power

Solution:

Let $m$ be the mass of the disc. Then translational kinetic energy of the disc is:
$K_{T}=\frac{1}{2}mv^{2}$ ....(1)
When it ascends on a smooth track its rotational kinetic energy will remain same while translational kinetic energy will go on decreasing. At highest point Translational kinetic energy is transferred to Potential Energy.
$K_{T}=mgh$
or $\frac{1}{2}mv^{2}=mgh$
or $h=\frac{v^{2}}{2 g}=\frac{\left(6\right)^{2}}{2 \times 10}=1.8 \, m$