Physics - Rotation Question with Solution | TestHub

The combination is rigid, so every part has $\omega =\frac{V}{4a}$ about the suspension point (disc centre is $3a+a=4a$ away). (A) KE of disc = translation of centre + spin: $\frac{1}{2}m{V}^2+\frac{1}{2}\frac{m{a}^2}{2}{\left(\frac{V}{4a}\right)}^2$ - correct. (B) rod 2's centre does not move with speed $V$ - wrong. (C) rod 1 turns about its end: $\frac{1}{2}\frac{m(3a{)}^2}{3}{\left(\frac{V}{4a}\right)}^2$ - correct. (D) total $KE=\frac{1}{2}\left[\frac{m(3a{)}^2}{3}+\left(\frac{m{a}^2}{2}+16m{a}^2\right)+\left(\frac{m(3a{)}^2}{12}+\frac{89m{a}^2}{4}\right)\right]{\left(\frac{V}{4a}\right)}^2$ - correct. (A), (C), (D)