A) 1:1
B) 4:1
C) 1:2
D) 2:1
Correct Answer: C
Solution :
M.I. of hollow cylinder is same as that of ring \[I=M{{R}^{2}}\] Kinetic energy when cylinder is sliding is \[{{E}_{1}}=\frac{1}{2}M{{\upsilon }^{2}}\] Kinetic energy when cylinder is rolling \[{{E}_{2}}=\frac{1}{2}M{{u}^{2}}+\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}M{{\upsilon }^{2}}+\frac{1}{2}M{{R}^{2}}\left( \frac{{{\upsilon }^{2}}}{{{R}^{2}}} \right)=M{{\upsilon }^{2}}\] so \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{1}{2}\]
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