A) \[\mu \,FR/2\]
B) \[\mu \,FR/3\]
C) \[\mu \,FR/6\]
D) \[\frac{2}{3}\mu \,FR\]
Correct Answer: D
Solution :
\[dN\,=\,\frac{F}{\pi {{R}^{2}}}\times 2\pi rdr\] \[dN\,=\,\frac{2Frdr}{{{R}^{2}}}\] \[{{f}_{k}}\,\,=\,\,\mu dN\] \[{{f}_{k}}\,=\frac{2Fr}{{{R}^{_{2}}}}\,\mu dr\] \[d{{\tau }_{f}}_{k}\,=\,{{f}_{k}}\,\times r\] \[{{d}_{\tau }}_{k}\,=\,\frac{2F\mu }{{{R}^{2}}}\,{{r}^{2}}dr\] \[{{_{\tau }}_{k}}\,=\,\int\limits_{0}^{R}{\frac{2F\mu }{{{R}^{2}}}\,{{r}^{2}}dr}\] \[=\,\,\frac{2F\mu }{{{R}^{2}}}\,\frac{{{R}^{3}}}{3}\,=\,\frac{2F\mu R}{3}\] Text will counter balance the torque by friction\[(\tau k)\] \[\therefore \,\,\,{{\tau }_{ext}}\,\,=\,\,\frac{2F\mu R}{3}\]You need to login to perform this action.
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