Question

A body moves on a horizontal circular road of radius $r$, with a tangential acceleration $a_1$. The coefficient of friction between the body and the road surface is $\mu$. The body begins to slip when its speed is $v$.

• A. $v=\mu rg$
• B. $\mu g=\frac{v^2}{r}+a_1$
• C. ${\mu }^2{g}^2=\frac{v^4}{r^2}+a_1^2$
• D. The force of friction makes an angle ${\tan }^{-1}\left(v^2/a_{1}r\right)$ with the direction of motion at the point of slipping

Answer 1

Answer: (C), (D)

The correct options are
C ${\mu }^2{g}^2=\frac{v^4}{r^2}+a_1^2$
D The force of friction makes an angle ${\tan }^{-1}\left(v^2/a_{1}r\right)$ with the direction of motion at the point of slipping

The tangential acceleration is $a_t=a_1$, and the radial acceleration $a_r=\frac{v^2}{r}$ being mutually perpendicular, have a resultant acceleration $a=\sqrt{a_t^2+a_r^2}$
The only horizontal force on he body is the force of friction, $F=mg=ma$, acting in the direction of $a$.
$\therefore \mu g=a\Rightarrow {\mu }^2{g}^2={\left(\frac{v^2}{r}\right)}^2+{a_1}^2=\frac{v^4}{r^2}+{a_1}^2$
Let the force of friction makes an angle $\theta$ with the direction of motion at the point of slipping.
$\therefore \mu mgsin\theta =\frac{v^2}{r}\&\mu mgcos\theta =a_1\Rightarrow tan\theta =\frac{v^2}{a_{1}r}\Rightarrow \theta ={tan}^{-1}\left(\frac{v^2}{a_{1}r}\right)$