Question

A small collar of mass $m$ is given an initial velocity of magnitude $v_o$ on the horizontal circular track fabricated from a slender rod. If the coefficient of kinetic friction is ${\mu }_k$, determine the distance travelled before the collar comes to rest. (Recognize that the frictional force depends on the net normal force).

• A. $\frac{r}{{\mu }_k}\ln \left[\frac{v_0^2+\sqrt{v_0^4+r^2{g}^2}}{rg}\right]$
• B. $\frac{r}{4{\mu }_k}\ln \left[\frac{v_0^2+\sqrt{v_0^4+r^2{g}^2}}{rg}\right]$
• C. $\frac{2r}{{\mu }_k}\ln \left[\frac{v_0^2+\sqrt{v_0^4+r^2{g}^2}}{rg}\right]$
• D. $\frac{r}{2{\mu }_k}\ln \left[\frac{v_0^2+\sqrt{v_0^4+r^2{g}^2}}{rg}\right]$

Answer 1

The correct option is D $\frac{r}{2{\mu }_k}\ln \left[\frac{v_0^2+\sqrt{v_0^4+r^2{g}^2}}{rg}\right]$

FBD of collar:


where $\frac{m{v}^2}{r}=$ centripetal force (acting towards centre),
$v=$ velocity of particle at any instant}
$N_x=\frac{m{v}^2}{r}$ and $N_y=mg$
So, $N=\sqrt{N_x^2+N_y^2}=\sqrt{{\left(mg\right)}^2+{\left(\frac{m{v}^2}{r}\right)}^2}$
$N=m\sqrt{g^2+\frac{v^4}{r^2}}$ ... (1)

Here, frictional force is acting opposite to the direction of motion. Hence, $f=-ma$
$f={\mu }_{k}N$ (${\mu }_k=$ kinetic friction coefficient)
So, ${\mu }_{k}N=-ma$
$\Rightarrow a=\frac{-{\mu }_{k}N}{m}$ ... (2)
From eq. (1)
$a=\frac{-{\mu }_k}{m}[m.\sqrt{g^2+\frac{v^4}{r^2}}]$ $[\because a=\frac{dv}{dt}]$
$\Rightarrow \frac{dv}{dx}.\frac{dx}{dt}=-{\mu }_k\sqrt{g^2+\frac{v^4}{r^2}}$ $[\because v=\frac{dx}{dt}]$
$\Rightarrow v.\frac{dv}{dx}=-{\mu }_k\sqrt{g^2+\frac{v^4}{r^2}}$

Let $\frac{v^2}{r}=k$
By differentiation, $\frac{2v}{r}.dv=dk$
or $v.dv=\frac{rdk}{2}$

So, $r.\frac{dk}{2dx}=-{\mu }_k\sqrt{g^2+k^2}$
Integrating both sides,
$\int \frac{r.dk}{2\sqrt{g^2+k^2}}=-{\mu }_k\int dx$
$\Rightarrow \frac{r}{2}\left[\ln [k+\sqrt{k^2+g^2}]\right]=-{\mu }_k\left[x\right]$
$\because (k=\frac{v^2}{r})$
$\Rightarrow \frac{r}{2}{\left(\ln [\frac{v^2}{r}+\sqrt{{\left(\frac{v^2}{r}\right)}^2+g^2}]\right)}_{v_o}^0=-{\mu }_k{\left[x\right]}_0^s$
[$\because$ initial velocity is $v_0$ and final velocity $=0$]

$\Rightarrow \frac{r}{2}[\ln \left(\sqrt{g^2}\right)-\left(\ln (\frac{v_0^2}{r}+\sqrt{{\left(\frac{v_0^2}{r}\right)}^2+g^2})\right)]=-{\mu }_{k}s$
$\Rightarrow s=\frac{r}{2{\mu }_k}\left[\ln \frac{v_0^2+\sqrt{v_0^4+g^2{r}^2}}{rg}\right]$