A) \[\frac{2m{{R}^{2}}\omega _{0}^{2}}{9}\]
B) \[\frac{m{{R}^{2}}\omega _{0}^{2}}{18}\]
C) \[\frac{m{{R}^{2}}\omega _{0}^{2}}{6}\]
D) \[\frac{m{{R}^{2}}\omega _{0}^{2}}{9}\]
Correct Answer: D
Solution :
| The angular speed of the disc when the balls reach the ends be\[\omega \]. From the law of conservation of angular momentum |
| \[\frac{1}{2}m{{R}^{2}}{{\omega }_{0}}=\frac{1}{2}m{{R}^{2}}\omega +\frac{m}{2}{{R}^{2}}\omega +\frac{m}{2}{{R}^{2}}\omega \] or \[\omega =\frac{{{\omega }_{0}}}{3}\] |
| The angular speed of the disc just after the balls leave the disc is \[\omega =\frac{{{\omega }_{0}}}{3}\] |
| Let the speed of each ball just after they leave the disc be v. |
| From principle of conservation of energy \[\frac{1}{2}\left( \frac{1}{2}m{{R}^{2}} \right)\omega _{0}^{2}=\frac{1}{2}\left( \frac{1}{2}m{{R}^{2}} \right){{\omega }^{2}}+\frac{1}{2}\left( \frac{m}{2} \right){{\text{v}}^{2}}+\frac{1}{2}\left( \frac{m}{2} \right){{\text{v}}^{2}}\] |
| By solving we get,\[\text{v=}\frac{2R{{\omega }_{0}}}{3}\] |
| Work done by all forces is equal to change in K.E., \[{{K}_{f}}-{{K}_{i}}=\frac{1}{2}\left( \frac{m}{2} \right){{\text{v}}^{2}}=\frac{m{{R}^{2}}\omega _{0}^{2}}{9}\] |
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