A) \[2a\frac{{{s}^{2}}}{R}\]
B) \[2as{{\left( 1+\frac{{{s}^{2}}}{{{R}^{2}}} \right)}^{1/2}}\]
C) \[2as\]
D) \[2a\frac{{{R}^{2}}}{s}\]
Correct Answer: B
Solution :
According to given problem \[\frac{1}{2}m{{v}^{2}}=a{{s}^{2}}\]\[\Rightarrow v=s\sqrt{\frac{2a}{m}}\] So \[{{a}_{R}}=\frac{{{v}^{2}}}{R}=\frac{2a{{s}^{2}}}{mR}\] ?(i) Further more as \[{{a}_{t}}=\frac{dv}{dt}=\frac{dv}{ds}\cdot \frac{ds}{dt}=v\frac{dv}{ds}\] ?(ii) (By chain rule) Which in light of equation (i) i.e. \[v=s\sqrt{\frac{2a}{m}}\] yields \[{{a}_{t}}=\left[ s\sqrt{\frac{2a}{m}} \right]\,\left[ \sqrt{\frac{2a}{m}} \right]=\frac{2as}{m}\] ?(iii) So that \[a=\sqrt{a_{R}^{2}+a_{t}^{2}}=\sqrt{{{\left[ \frac{2a{{s}^{2}}}{mR} \right]}^{2}}+{{\left[ \frac{2as}{m} \right]}^{2}}}\] Hence \[a=\frac{2as}{m}\sqrt{1+{{[s/R]}^{2}}}\] \ \[F=ma=2as\sqrt{1+{{[s/R]}^{2}}}\]
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