Question

a circular disc of mass 0.5kg and radius 0.1m is making 60 revolutions per n=minute about an axis passing through the centre and perpendicular to its plane . calculate it's kinetic energy​

Answer 1

$\huge\mathcal\colorbox{lavender}{{\color{b}{✿answer✿:}}}$

${{given:}} \\ \red{mass \: of \: disc \: (m) = 0.5kg} \\ \red{radius \: of \: disc \: (r) = 0.1m} \\ \red {angular \: velocity \: of \: disc(w) =60rpm = \frac{60 \times 2\pi }{60} = 2\pi \: rad/sec} \\ \blue{moment \: of \: intertia \: of \: disc \: about \: a \: point \: passing \: through} \\ \blue{circumference \: and \: perpendicular \: to \: the \: plane}\\ \green{{I}^{'} = Ig + m{r}^{2}} \\ \pink{but \: \: Ig = \frac{ m{r}^{2} }{2} (for \: above \: case)} \\ \green{∴{I}^{'} = \frac{m{r}^{2} }{2} + m{r}^{2} = \frac{3}{2} m {r}^{2}} \\ \orange{Rotational \: kinetic \: energy  \: = \frac{1}{2} I{ω}^{2}} \\   \orange{ = \frac{1}{2} \times \frac{3}{2} M{R}^{2} ×{(2π)}^{2}}\\ \blue{∴Rotational \: kinetic \: energy = \frac{3}{4} \times 0.5 \times {(0.1)}^{2} \times {(2 \times 3.142)}^{2}}\\ = \green{0.00375 \times 2 \times 3.142 \times 2 \times 3.142 }\\ = \green{0.14808246}$