Question

Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiments, they studied the maximum speed that quolls could run around a curved path without slipping. One quoll was running at 3.2 m/s around a curve with a radius of 1.4 m when it started to slip.What was the coefficient of static friction between the quoll's feet and the ground in this trial?

Answer 1

Answer:

$\mu = 0.746$

Explanation:

As we know that the force required to move in circle at uniform speed is known as centripetal force

So here we know that centripetal force is given as

$F = \frac{mv^2}{R}$

so here the force is due to friction force

so it is given as

$\mu mg = \frac{mv^2}{R}$

$\mu = \frac{v^2}{Rg}$

now we have

$\mu = \frac{3.2^2}{1.4 \times 9.8}$

$\mu = 0.746$

Answer 2

Answer:

$\mu=0.74$            

Explanation:

It is given that,

Speed of one quoll around a curve, v = 3.2 m/s (maximum speed)

Radius of the curve, r = 1.4 m

On the curve, the centripetal force is balanced by the frictional force such that the coefficient of frictional is given by :

$\mu=\dfrac{v^2}{rg}$

$\mu=\dfrac{(3.2\ m/s)^2}{1.4\ m\times 9.8\ m/s^2}$

$\mu=0.74$

So, the coefficient of static friction between the quoll's feet and the ground in this trial is 0.74. Hence, this is the required solution.