A) 1.2 m/s
B) 2\[\sqrt{2}\] m/s
C) 20 m/s
D) 2 m/s
Correct Answer: B
Solution :
Rotational kinetic energy of the thin metal \[disc\,{{E}_{K(rot)}}=\frac{1}{2}I{{\omega }^{2}}\]and \[I=\frac{m{{R}^{2}}}{2}\] \[{{E}_{K\left( rot \right)}}=\frac{1}{2}I\frac{{{v}^{2}}}{{{r}^{2}}}\] \[{{E}_{K(rot)}}=\frac{1}{4}m{{r}^{2}}\times \frac{{{v}^{2}}}{{{r}^{2}}}\] \[4=\frac{1}{4}m{{v}^{2}}\] \[{{v}^{2}}=\frac{4\times 4}{m}=\frac{4\times 4}{2}=8\] \[v=\sqrt{8}=2\sqrt{2}\,m/s\]You need to login to perform this action.
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