Question

The kinetic energy of a circular disc rotating with a speed of $60$ r.p.m. about an axis passing through a point on its circumference and perpendicular to its plane is (mass of circular disc = $5kg$, radius of disc = $1m$) approximately:

• A. $170J$
• B. $-160J$
• C. $150J$
• D. $140J$

Answer 1

Answer: (C)

$150J$

$KE=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times \frac{3}{2}m{r}^{2}4{\pi }^2{f}^2$$=3m{r}^2\times {\pi }^2{f}^2=3\times 5\times 1\times 10\times 1$

$KE=150J$