Question

A uniform circular disc of mass m is set rolling on a smooth horizontal table with a uniform linear velocity v. Find the total kinetic energy of the disc?
1. \(\frac{1}{2}m{v^2}\)
2. \(\frac{1}{4}m{v^2}\)
3. \(\frac{3}{4}m{v^2}\)
4. None

Answer 1

Correct Answer - Option 3 : \(\frac{3}{4}m{v^2}\)

MI of the disc about its own axis, \(I = \frac{1}{2}m{r^2}\)

As \(v = r\omega \Rightarrow {\omega ^2} = \frac{{{v^2}}}{{{r^2}}}\)

\(\begin{array}{l} Total\;kinetic\;energy = Rotational\;KE + Translational\;KE\\ Total\;kinetic\;energy = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2}\\ \Rightarrow \frac{1}{2} \times \frac{1}{2}m{r^2} \times \frac{{{v^2}}}{{{r^2}}} + \frac{1}{2}m{v^2} = \frac{3}{4}m{v^2} \end{array}\)