Question 297: Work, Energy and Power & Circular Motion

Question

A child starts running from rest along a circular track of radius $r$ 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
[g = acceleration due to gravity]

Accepted Answer

When child moves in a circular track, he is acted upon by two force as shown below,
Here, $f_c=f \sin 	heta$ and $f_t=f \cos 	heta$.
As, $f_c$ is the centripetal force and $f_t$ is the tangential force. So,
$$\begin{aligned}
& f_c=\frac{m v^2}{r}=f \sin 	heta \
	ext{and }\quad& f_t=m a=f \cos 	heta
\end{aligned}$$
$	herefore$ Resultant force, $f_R=\sqrt{(f \sin 	heta)^2+(f \cos 	heta)^2}$
$$=\sqrt{\left(\frac{m v^2}{r}ight)^2+(m a)^2}$$
Also, when the shoes starts slipping, the friction becomes equal to resultant force.
$$\begin{array}{lc}
	herefore & f=f_R \
\Rightarrow & \mu m g=\sqrt{\left(\frac{m v^2}{r}ight)^2+(m a)^2} \
\Rightarrow & \mu^2 m^2 g^2=\frac{m^2 v^4}{r^2}+m^2 a^2 \
\Rightarrow& \mu^2 g^2=\frac{(a t)^4+a^2 r^2}{r^2} \quad\left(\because a=\frac{v}{t}ight) \
	ext { or } & \mu=\frac{\left(a^4 t^4+a^2 r^2ight)^{\frac{1}{2}}}{g r}
\end{array}$$

Answer

$\frac{\left[a^4 t^4+a^2 r^2\right]^{\frac{1}{2}}}{g r}$

Answer

$\frac{\left[a^4 t^4+a^2 r^2\right]}{r g}$

Answer

$\frac{\left[a^2 t^2+a^4 r^4\right]}{r g}$

Answer

$\frac{\left[a^4 t^4-a^2 r^2\right]^{\frac{1}{2}}}{r g}$

Work, Energy and Power & Circular Motion

A child starts running from rest along a circular track of radius r 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 [g = acceleration due to gravity]

Choose the correct answer: