Question

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

Hint:

For a rolling object, the total kinetic energy is the sum of translational and rotational kinetic energy, and the relationship between speed and moment of inertia depends on the object’s geometry.

Answer 1

Correct Answer: 10

To solve the problem of finding the speed of the center of mass of a rolling solid sphere, we start by considering the kinetic energy of the sphere. The total kinetic energy \( KE \) of a rolling object is the sum of its translational and rotational kinetic energy:

\( KE = KE_{\text{trans}} + KE_{\text{rot}} \)

For a solid sphere rolling without slipping, the translational kinetic energy is given by:

\( KE_{\text{trans}} = \frac{1}{2} mv^2 \)

And the rotational kinetic energy is given by:

\( KE_{\text{rot}} = \frac{1}{2} I \omega^2 \)

For a solid sphere, the moment of inertia \( I \) is \( \frac{2}{5}mr^2 \) and the angular velocity \( \omega \) is related to the linear velocity \( v \) by \( \omega = \frac{v}{r} \). Hence, the rotational kinetic energy becomes:

\( KE_{\text{rot}} = \frac{1}{2} \times \frac{2}{5}mr^2 \left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2 \)

Substitute the expressions for \( KE_{\text{trans}} \) and \( KE_{\text{rot}} \) into the equation for total kinetic energy:

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

We are given \( KE = 7 \times 10^{-3} \, J \) and \( m = 1 \, kg \). Substituting these values, we get:

\( 7 \times 10^{-3} = \frac{7}{10} \times 1 \times v^2 \)

Simplifying,

\( v^2 = 10 \times 7 \times 10^{-3} / 7 = 10 \times 10^{-3} \)

\( v^2 = 10^{-2} \)

\( v = \sqrt{10^{-2}} = 0.1 \, m/s \)

Converting this speed to cm/s:

\( v = 0.1 \times 100 = 10 \, cm/s \)

The speed of the center of mass of the sphere is \( 10 \, cm/s \), which fits within the provided range of 10,10.