A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of the center of mass of the sphere will be –––– ms^-1.

Given: mass of the solid sphere m = 2 kg, total kinetic energy KE = 2240 J, and the sphere is in pure rolling. Find: the velocity of the center of mass v. For a rolling solid sphere, total kinetic energy is KE = 12mv^2 + 12I ^2 where for a solid sphere, I = 25mr^2 and for pure rolling, = vr Substitute these into the kinetic energy expression: KE = 12mv^2 + 12 ( 25mr^2) ( v^2r^2) So, KE = 12mv^2 + 15mv^2 = 710mv^2 Now substitute the given values: 2240 = 710 2 v^2 v^2 = 1600 v = 40 ms^-1 Therefore, the velocity of the center of mass is 40 ms^-1.

A solid sphere of mass 2 kg is making pure rolling on a horizonta… — Solution | JEE Physics

NVAEasyJEE 2023Rolling Motion & Rotational Kinematics

A solid sphere of mass $2 \, \text{kg}$ is making pure rolling on a horizontal surface with kinetic energy $2240 \, \text{J}$ . The velocity of the center of mass of the sphere will be –––– $\text{ms}^{-1}$ .

Answer

Correct answer:40

Step-by-step solution

Standard Method

Given: mass of the solid sphere $m = 2 \, \text{kg}$ , total kinetic energy $KE = 2240 \, \text{J}$ , and the sphere is in pure rolling.

Find: the velocity of the center of mass $v$ .

For a rolling solid sphere, total kinetic energy is

KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

where for a solid sphere,

I = \frac{2}{5}mr^2

and for pure rolling,

\omega = \frac{v}{r}

Substitute these into the kinetic energy expression:

KE = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v^2}{r^2}\right)

So,

KE = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2

Now substitute the given values:

2240 = \frac{7}{10} \times 2 \times v^2

v^2 = 1600

v = 40 \, \text{ms}^{-1}

Therefore, the velocity of the center of mass is $40 \, \text{ms}^{-1}$ .

Common mistakes

Using only translational kinetic energy $\frac{1}{2}mv^2$ and ignoring rotational kinetic energy is wrong because the sphere is in pure rolling. Include both translational and rotational parts in the total kinetic energy.
Using the wrong moment of inertia is incorrect. For a solid sphere, $I = \frac{2}{5}mr^2$ , not the expression for a ring, disc, or hollow sphere.
Forgetting the pure rolling condition $\omega = \frac{v}{r}$ leads to an incomplete substitution. Replace $\omega$ in terms of $v$ before simplifying the energy expression.

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