A) (a )\[m{{\upsilon }^{2}}\]
B) \[\frac{3}{4}m{{\upsilon }^{2}}\]
C) \[\frac{1}{4}m{{\upsilon }^{2}}\]
D) \[\frac{1}{2}m{{\upsilon }^{2}}\]
Correct Answer: B
Solution :
Total kinetic energy of circular disc when it is rolling down is \[={{(K.E.)}_{\text{translational}}}+{{(K.E.)}_{\text{rotational}}}\] \[=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}\frac{m{{r}^{2}}}{2}.\frac{{{\upsilon }^{2}}}{{{r}^{2}}}\] \[=\frac{3}{4}m{{\upsilon }^{2}}\]
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