Saved Bookmarks
| 1. |
Calculate the M.I. and rotational K.E. of a thin uniform rod of mass \( 10 g \) and length \( 60 cm \), wher it rotates about a transverse axis through its centre at \( 90 rpm \) |
|
Answer» Given L = 60 cm L = 0.60 m M = 10 g M = .010 kg n = 90 rpm n = \(\frac{90}{60}\) = 1.5 rps We know that, moment of inertia thin uniform rod transverse axis through its centre. I = \(\frac{ML^2}{12}\) I = \(\frac{.010 \times (0.60)^2}{12}\) I = 0.0003 I = 3 × 10-4 kg m2 Rotational kinetic energy K.E = \(\frac{1}{2}\,I\omega^2\) ∵ \(\omega = 2\pi n\) \(=\frac{1}{2} \times (0.0003) \times (2\pi n)^2\) \(=\frac{1}{2} \times (0.003) \times (2\pi \times 1.5)^2\) \(=\frac{1}{2} \times 0.003 \times 9\pi^2\) K.E = 0.133 Joule |
|