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

\(\frac1{2}mv^2 +\frac1{2}m{ω}^2 = 7 \times 10^{-3}\)

\(\frac1{2}mv^2 +\frac1{2} (\frac2{5}MR^2)({\frac{V}{R}})^2 = 7 \times 10^{-3}\)

\(\frac1{2}MV^2 +[1+\frac2{5}]  = 7 \times 10^{-3}\)

\(\frac1{2}(1)V^2 +\frac7{5}  = 7 \times 10^{-3}\)

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

\(V = 10^{-1}\)

\(V= 0.1\,ms^{-1} = 10\,cm^{-1}\)

So, The correct answer is 10.