Question

A child starts running from rest along a circular track of radius r with constant tangential acceleration a. After time the feels that slipping of shoes on the ground has started. The coefficient of friction between shoes and the ground is [g = acceleration due to gravity].

विकल्प

• ${\displaystyle \frac{{[{\text{a}}^4{\text{t}}^4+{\text{a}}^2{\text{r}}^2]}^{\frac{1}{2}}}{\text{gr}}}$

• ${\displaystyle \frac{[{\text{a}}^4{\text{t}}^4+{\text{a}}^2{\text{r}}^2]}{\text{rg}}}$

• ${\displaystyle \frac{[{\text{a}}^2{\text{t}}^2+{\text{a}}^4{\text{r}}^4]}{\text{rg}}}$

• ${\displaystyle \frac{{[{\text{a}}^4{\text{t}}^4-{\text{a}}^2{\text{r}}^2]}^{\frac{1}{2}}}{\text{rg}}}$

Answer 1

${\displaystyle \frac{{[{\text{a}}^4{\text{t}}^4+{\text{a}}^2{\text{r}}^2]}^{\frac{1}{2}}}{\text{gr}}}$

Explanation:

When child moves in a circular track, he is acted upon by two force as shown below.

Here, f_c = f sin Sand f_t = f cos θ.
As, f_c is the centripetal force and f_t is the tangential force. So,

${\displaystyle {\text{f}}_{\text{c}}=\frac{{\text{mv}}^2}{\text{r}}}$= f cos θ

and f_t = ma = f cos θ

∴ Resultant force, f_R = ${\displaystyle \sqrt{{\left(\text{f}\sin \theta \right)}^2+{\left(\text{f}\cos \theta \right)}^2}}$

${\displaystyle =\sqrt{{\left(\frac{{\text{mv}}^2}{\text{r}}\right)}^2+{\left(\text{ma}\right)}^2}}$

Also, when the shoes starts slipping, the friction becomes equal to resultant force.

∴ f = f_R

${\displaystyle \mu \text{mg}=\sqrt{{\left(\frac{{\text{mv}}^2}{\text{r}}\right)}^2+{\left(\text{ma}\right)}^2}}$

${\displaystyle \Rightarrow {\mu }^2{\text{m}}^2{\text{g}}^2=\frac{{\text{m}}^2{\text{v}}^4}{{\text{r}}^2}+{\text{m}}^2{\text{a}}^2}$

${\displaystyle \Rightarrow {\mu }^2{\text{g}}^2=\frac{{\left(\text{at}\right)}^4+{\text{a}}^2{\text{r}}^2}{{\text{r}}^2} ....(\because \text{a}=\frac{\text{v}}{\text{t}})}$

or ${\displaystyle \mu =\frac{{[{\text{a}}^4{\text{t}}^4+{\text{a}}^2{\text{r}}^2]}^{\frac{1}{2}}}{\text{gr}}}$