A child starts running from rest along a circular track of radius ' $\mathrm{r}^{\prime}$ ' with constant…

A child starts running from rest along a circular track of radius ' $\mathrm{r}^{\prime}$ ' with constant tangential acceleration 'a'. After time 't' he feels that slipping of shoes on the ground has started. The coefficient of friction between shoes and the ground is $[\mathrm{g}=$ acceleration due to gravity $]$
  1. $\frac{\left[a^{2} t^{2}+a^{4} r^{4}\right]}{r g}$
  2. $\frac{\left[a^{4} t^{4}+a^{2} r^{2}\right]}{r g}$
  3. $\frac{\left[a^{4} t^{4}+a^{2} r^{2}\right]^{\frac{1}{2}}}{g r}$
  4. $\frac{\left[a^{4} t^{4}-a^{2} r^{2}\right]^{\frac{1}{2}}}{r g}$

Solution

After time t, velocity $\mathrm{V}=$ at $\therefore$ radial acceleration $a_{\mathrm{r}}=\frac{V^{2}}{r}=\frac{a^{2} t^{2}}{r}$ Total acceleration $=\sqrt{\frac{a^{4} t^{4}}{r^{2}}+a^{2}}$ $\therefore \mu g=\sqrt{\frac{a^{4} t^{4}+a^{2} r^{2}}{r^{2}}}$ $\therefore \mu=\frac{\left[a^{4} t^{4}+a^{2} r^{2}\right]^{\frac{1}{2}}}{g r}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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