Question

A motorcyclist of mass $m$ is to negotiate a curve of radius $r$ with a speed $v$. The minimum value of the coefficient of friction so that this negotiation may take place safely is?
A. $v^{2}rg$
B. ${\displaystyle \frac{v^2}{gr}}$
C. ${\displaystyle \frac{gr}{v^2}}$
D. ${\displaystyle \frac{g}{v^{2}r}}$

Answer 1

Hint: Since there is a downward force (which equals the weight of the body) applied to a body. When a body takes a curve with radius $r$ there is a chance of slipping a body hence, to negotiate the slip the normal force acting opposite to the downward force must be balanced by the centrifugal force.

Complete answer:
Mass of a motorcyclist $=m$ (given)
Since the gravity $g$ acts downward. Therefore, the weight of a body $=mg$
Let us consider the normal force $N$ is equal to and opposite to the direction of weight$(mg)$ of a body.

Now, we know that Centrifugal Force acting on the body:
$F_c=m{\displaystyle \frac{v^2}{r}}$ where,
v = speed of a body
r = radius of curve taken by body

Also, we know that $F=\mu N$where,
F = Frictional Force and$\mu$= Coefficient of friction

To avoid slip, the frictional force must be balanced by centrifugal force i.e.,
$F=F_c$
$\mu N=m{\displaystyle \frac{v^2}{r}}$

Substitute $N=mg$ in the above expression, we get
$\mu (mg)=m{\displaystyle \frac{v^2}{r}}$
$\mu ={\displaystyle \frac{v^2}{gr}}$

Thus, the minimum value of the coefficient of friction so that this negotiation may take place safely is $\mu ={\displaystyle \frac{v^2}{gr}}$.

Hence, the correct option is (B) $\mu ={\displaystyle \frac{v^2}{gr}}$ >

Note: Since this is a problem based on the balancing of two different forces hence, given conditions are to be analyzed very carefully and only after which the procedure of solving the problem is identified. To have a better understanding of the formulas used, it is essential to understand which kind of forces influences the problem.