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NEET 2018
Exam
NEET
Year
2018
Subject
Physics
Chapter
Rotational Motion

NEET 2018 · Physics · Rotational Motion

A solid sphere is in rolling motion. In rolling motion a body...

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (Kt)\left({K}_{t}\right) as well as rotational kinetic energy \left({K}_{r}\right)simultaneously.Theratiosimultaneously. The ratio{K}_{t}: \left({K}_{t} + {K}_{r}\right) for the sphere is

Options

  1. A10 : 7
  2. B5 : 7
  3. C7 : 10
  4. D2 : 5

Correct answer

B. 5 : 7

Step-by-step solution

Transnational KEKE =12mv2= \dfrac{1}{2} m{v}^{2}
Translational KE+KE + Rotational KEKE
=12mv2+12Iω2= \dfrac{1}{2} m{v}^{2} + \dfrac{1}{2} I{\omega}^{2}
=12mv2+12(25mr2)ω2= \dfrac{1}{2} m{v}^{2} + \dfrac{1}{2} ( \dfrac{2}{5} m{r}^{2} ){\omega}^{2}
∵∵ for rolling motion v=rωv = r\omega
=12mv2+12×25mr2(vr)2=710mv2= \dfrac{1}{2} m{v}^{2} + \dfrac{1}{2} \times \dfrac{2}{5} m{r}^{2} {( \dfrac{v}{r} )}^{2} = \dfrac{7}{10} m{v}^{2}
So, KtKt+Kr=12mv2710mv2=57\dfrac{{K}_{t}}{{K}_{t} + {K}_{r}} = \dfrac{\dfrac{1}{2} m{v}^{2}}{\dfrac{7}{10} m{v}^{2}} = \dfrac{5}{7}
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