A) 121.4 J
B) 185J
C) 15.2 J
D) 22.0 J
Correct Answer: A
Solution :
Moment of inertia about diameter is \[{{I}_{G}}=\frac{2}{5}M{{r}^{2}}\] Distance of diameter from rotational axis \[d=1\text{ }m+10\text{ }cm=1.1\text{ }m\] From the theorem of parallel axis \[I={{I}_{G}}+M{{r}^{2}}\] \[=\frac{2}{5}M{{r}^{2}}+M{{d}^{2}}\] \[=\frac{2}{5}\times 2\times {{(0.1)}^{2}}+2\times {{(1.1)}^{2}}\] \[=0.008+2.42\] \[=2.428\text{ }kg-{{m}^{3}}\] Therefore, rotational kinetic energy of sphere \[K=\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}\times 2.428\times {{(10)}^{2}}\] \[=121.4\text{ }J\]You need to login to perform this action.
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