Physics - Rotation Question with Solution | TestHub

Mass of the hoop, $m=100$ kg

Speed of centre of mass, $v=20$ cm/s $=0.2$ m/s

Total kinetic energy of a rolling hoop is the sum of translational and rotational kinetic energies.

$K=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

For a hoop (thin ring), moment of inertia $I=m{R}^2$ and for rolling without slipping, $\omega =\frac{v}{R}$.

$K=\frac{1}{2}m{v}^2+\frac{1}{2}(m{R}^2){\left(\frac{v}{R}\right)}^2$

$K=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2=m{v}^2$

Substituting the given values:

$K=100\times (0.2{)}^2$

$K=100\times 0.04=4$ J

Answer: $4$ J.