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Rotational Motion question

2023 · 31 Jan · Shift 1 · Q69
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  5. /2023 · 31 Jan · Shift 1 · Q69

Rotational Motion question

2023 · 31 Jan · Shift 1 · Q69

JEE MainPhysicsRotational MotionNumerical+4 / −1
A solid sphere of mass 1 kg1 \mathrm{~kg}1 kg rolls without slipping on a plane surface. Its kinetic energy is 7×10−3 J7 \times 10^{-3} \mathrm{~J}7×10−3 J. The speed of the centre of mass of the sphere is ‾cm s⁡−1\underline{\hspace{2cm}}\operatorname{cm~s}^{-1}​cm s−1
Numerical answer
View written solutionFree

Correct answer: 10

  1. Write total kinetic energy for rolling without slipping

For a body rolling without slipping, K=12mv2+12Iω2K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2K=21​mv2+21​Iω2 with v=ωR.v = \omega R.v=ωR.

  1. Moment of inertia of a solid sphere

For a solid sphere about its centre, I=25mR2.I = \frac{2}{5}mR^2.I=52​mR2.

So rotational kinetic energy becomes

= \frac{1}{5}mv^2.$$ 3. **Total kinetic energy** Hence, $$K = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \left(\frac{5}{10}+\frac{2}{10}\right)mv^2 = \frac{7}{10}mv^2.$$ 4. **Substitute given values** Given: $$m = 1\text{ kg}, \qquad K = 7\times 10^{-3}\text{ J}.$$ So, $$7\times 10^{-3} = \frac{7}{10}(1)v^2.$$ Cancel 7 on both sides: $$10^{-3} = \frac{1}{10}v^2$$ $$v^2 = 10^{-2}$$ $$v = 10^{-1}\text{ m/s} = 0.1\text{ m/s}.$$ 5. **Convert to cm/s** $$0.1\text{ m/s} = 10\text{ cm/s}.$$ ## Final Answer $$\boxed{10}$$
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