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NVAEasyJEE 2023Rolling Motion & Rotational Kinematics

JEE Physics 2023 Question with Solution

A solid sphere of mass 1 kg1 \, \text{kg} rolls without slipping on a plane surface. Its kinetic energy is 7×10−3 J7 \times 10^{-3} \, \text{J}. The speed of the centre of mass of the sphere is ..... cm/s.

Answer

Correct answer:10

Step-by-step solution

Standard Method

Given: A solid sphere of mass 1 kg1 \, \text{kg} rolls without slipping on a plane surface. Its total kinetic energy is 7×10−3 J7 \times 10^{-3} \, \text{J}.

Find: The speed of the centre of mass in cm/s.

For a rolling object, total kinetic energy is the sum of translational and rotational kinetic energies:

12mv2+12Iω2=7×10−3\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = 7 \times 10^{-3}

For a solid sphere,

I=25MR2I = \frac{2}{5}MR^2

and for rolling without slipping,

ω=VR\omega = \frac{V}{R}

Substituting these into the kinetic energy expression,

12mv2+12(25MR2)(VR)2=7×10−3\frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\left(\frac{V}{R}\right)^2 = 7 \times 10^{-3}

This gives,

12MV2+15MV2=7×10−3\frac{1}{2}MV^2 + \frac{1}{5}MV^2 = 7 \times 10^{-3}

so,

710MV2=7×10−3\frac{7}{10}MV^2 = 7 \times 10^{-3}

With M=1M = 1,

710V2=7×10−3\frac{7}{10}V^2 = 7 \times 10^{-3}

Therefore,

V2=10−2V^2 = 10^{-2}

and hence,

V=10−1 m/sV = 10^{-1} \, \text{m/s}

Converting to cm/s,

V=0.1 m/s=10 cm/sV = 0.1 \, \text{m/s} = 10 \, \text{cm/s}

Therefore, the speed of the centre of mass is 10 cm/s10 \, \text{cm/s}.

Using rolling kinetic energy formula

Given: A solid sphere rolls without slipping and has total kinetic energy 7×10−3 J7 \times 10^{-3} \, \text{J}.

Find: The speed of its centre of mass.

For a rolling solid sphere,

K=12Mv2+12Iω2K = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2

with

I=25MR2,ω=vRI = \frac{2}{5}MR^2, \qquad \omega = \frac{v}{R}

Hence,

K=12Mv2+12(25MR2)(vR)2K = \frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\left(\frac{v}{R}\right)^2 =12Mv2+15Mv2= \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 =710Mv2= \frac{7}{10}Mv^2

Now substitute K=7×10−3 JK = 7 \times 10^{-3} \, \text{J} and M=1 kgM = 1 \, \text{kg}:

710v2=7×10−3\frac{7}{10}v^2 = 7 \times 10^{-3} v2=10−2v^2 = 10^{-2} v=10−1 m/sv = 10^{-1} \, \text{m/s}

Finally,

10−1 m/s=10 cm/s10^{-1} \, \text{m/s} = 10 \, \text{cm/s}

So, the correct numerical answer is 10.

Common mistakes

  • Using only translational kinetic energy 12mv2\frac{1}{2}mv^2 is incorrect because the sphere is rolling without slipping, so rotational kinetic energy must also be included. Use total kinetic energy as the sum of translational and rotational parts.

  • Using the wrong moment of inertia is a common error. For a solid sphere, the correct value is I=25MR2I = \frac{2}{5}MR^2, not the expressions for a ring, disc, or hollow sphere.

  • Forgetting the rolling condition ω=vR\omega = \frac{v}{R} leads to an incomplete substitution. Replace angular speed using the no-slip condition before simplifying the energy equation.

  • Stopping at 0.1 m/s0.1 \, \text{m/s} and entering that as the answer is wrong because the question asks for cm/s. Convert properly: 0.1 m/s=10 cm/s0.1 \, \text{m/s} = 10 \, \text{cm/s}.

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