Question

A particle of mass $m$ attached to a string, describes a horizontal circle of radius $r$ on a rough table at speed $v_0$. After completing one full trip around the circle, the speed of the particle is halved. If the coefficient of friction is $\frac{\left(3v_0^2\right)}{\left(a\pi gr\right)}$, then find $a$.

Answer 1

Given
$\mu =\frac{\left(3v_0^2\right)}{\left(a\pi gr\right)}$
Initial velocity, $v_i=v_o$
Final velocity, $v_f=\frac{v_0}{2}$
Work done by friction force $=-\mu mg2\pi r$
Form the work energy conservation,
$W_f=\Delta K$
$=\mu mg2\pi r=\frac{1}{2}m\left(\frac{v_0}{2}\right)-\frac{1}{2}m{v}_0^2$
substitute the value, we get
$\frac{3v_0^2}{a\pi gr}\cdot mg\cdot 2\pi r=-\frac{3m{v}_0^2}{8}$
$a=16$
Final answer: $16$