A) \[\frac{3}{4}M{{\upsilon }^{2}}\]
B) \[\frac{2}{5}M{{\upsilon }^{2}}\]
C) \[\frac{3}{2}M{{\upsilon }^{2}}\]
D) \[1\frac{1}{3}M{{\upsilon }^{2}}\]
Correct Answer: A
Solution :
The total energy of a body rolling without slipping \[{{K}_{total}}={{K}_{rot}}+{{K}_{trans}}\] \[=\frac{1}{2}I{{\omega }^{2}}+\frac{1}{2}M{{\upsilon }^{2}}\] But \[I\,disc=\frac{1}{2}M{{r}^{2}}\,and\,\upsilon =r\omega \] \[\therefore \] \[{{K}_{total}}=\frac{1}{2}\left( \frac{1}{2}M{{r}^{2}} \right){{\omega }^{2}}+\frac{1}{2}M{{(r\omega )}^{2}}\] \[=\frac{1}{4}M{{r}^{2}}{{\omega }^{2}}+\frac{1}{2}M\,{{r}^{2}}{{\omega }^{2}}\] \[=\frac{3}{4}M{{r}^{2}}{{\omega }^{2}}\] \[=\frac{3}{4}M{{\upsilon }^{2}}\]
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